Coordinate Geometry

3.44 Coordinate Geometry Booster

56. Let the chord AB is bisected at C(x1, y1) 61. Obviously, the locus of the mid-point of the chords of
the circle is the diameter of the given circle.
The equation of AB is xx1 + yy1 = x12 + y12 Here, the centre of the circle x2 + y2 – 4x – 6y = 0 is
(2, 3).
which is passing through (h, k), so The equation of the diameter of the circle, which is per-
hx1 + ky1 = x12 + y12 pendicular to the line 4x + 5y + 10 = 0 is 5x – 4y + k = 0
which is passing through (2, 3).
Thus, the locus of (x1, y1) is Therefore, k = 12 – 10 = 2.
x2 + y2 = hx + ky Hence, the equation of the diameter is
5x – 4y + 2 = 0
57. Let M(h, k) is the mid-point of AB
62. The equation of the common chord of the circle is
In DOAM, 2x – 2y = 0

sin (45°) = OM = OM O fi x=y
AM 2 M On solving y = x and the circle

fi OM = 2 sin (45°) A B x2 + y2 + 3x + 5y + 4 = 0,
we get the points of intersection.
=2¥ 1 = 2
2 Thus, x = - 2 + 2 = y and x = - 2 - 2 = y

fi OM 2 = 2 Let the common chord be PQ, where
fi h2 + k2 = 2
Hence, the locus of (h, k) is x2 + y2 = 2.

58. Let A represents a complex number Z. P = (- 2 + 2, - 2 + 2) and

i.e. Z = 3 + i Q = (-2 - 2, -2 - 2)
Here, –AOB = 90°.
Thus, the length of the common chord, PQ

By rotation theorem, we can say that = (- 2 - 2 + 2 - 2)2 + (- 2 - 2 + 2 - 2)2

B = iZ = i ( 3 + i) = –1 + i 3

Hence, the point B is (-1, 3). = 8 + 8 = 16 = 4

59. Let C represents a complex number Z and D be Z1. 63. The equation of the common chord of the given circles
Here, –COD = 60° is 2x + 1 = 0.
On solving 2x + 1 = 0 and
By rotation theorem, Z1 - 0 = Z1 - 0 ¥ e - i p x2 + y2 + 2x + 3y + 1 = 0,
3 we get the points of intersection.
Thus, the points of intersection are
Z -0 Z -0
x = - 1 and y = - 3 ± 2
fi Z1 = |Z1| ¥ e-i p = e-i p 22
3 3

Z |Z|

fi Z1 = Ze - i p
3

fi Z1 = (2 2 + Ê 1 - i 3ˆ
i) ËÁ 2 2 ˜¯
Let PQ be the common chord, where

fi Ê 2+ 3ˆ + i Ê 1 - 3¯˜ˆ P = Ê — 1 , - 3 + 2˜¯ˆ and
Z1 = ÁË 2 ˜¯ ÁË 2 ËÁ 2 2

Hence, the co-ordinates of D are Q Ê — 1 , 3 2 ¯ˆ˜
ËÁ 2 2
Ê 2+ 3,1 - ˆ = - -
ËÁ 22 3˜¯
The mid-point of P and Q is the centre of the new cir-

60. As we know that the locus of the mid-points of the cle.
chord of the circle is the diameter of the circle. Thus,
the equation of the diameter which is parallel to Thus, centre is C = Ê — 1 , - 3ˆ
2012x + 2013y + 2014 = 0 ËÁ 2 2¯˜
is 2012x + 2013y + k = 0
which is passing through the centre of the circle x2 + y2 Now, radius r = CP = 2 .
= 9.
Therefore, k = 0 Hence, the equation of the circle is
Hence, the equation of the diameter is
2012x + 2013y = 0 Ê 1ˆ 2 Ê 3ˆ 2
ÁË 2¯˜ ËÁ 2˜¯
x + + y + = 2

fi 2(x2 + y2) + 2x + 6y + 1 = 0

Circle 3.45

64. (i) The given circles are x2 + y2 = 4 …(i) (vi) The given circles are
and x2 + y2 – 2x = 0 (ii) x2 + y2 - 2(1 + 2) x - 2y + 1 = 0

Now, C1(0, 0) and C2(1, 0) and r1 = 2 and r2 = 1. and x2 + y2 – 2x – 2y + 1 = 0
Thus, C1C2 = 1 = r2 – r1
Hence, both the circles touch each other internally. Now C1: (1 + 2, 1) , C2: (1, 1),

Thus, the number of common tangent = 1. r1 = 2 + 1 and r2 = 1.
(ii) The given circles are x2 + y2 + 4x + 6y + 12 = 0
Therefore, C1C2 = ( 2)2 = 2
and x2 + y2 – 6x – 4y + 12 = 0
and r1 – r2 = 2 .
Now, C1: (–2, –3), C2: (3, 2), r1 = 1 and r2 = 1 Thus,C1C2 = r1 – r2
Therefore, Hence, the number of common tangents = 1.

C1C2 = (3 + 2)2 + (- 3 - 2)2 65. The given circles are
= 25 + 25 = 50 = 5 2 x2 + y2 – 2x – 6y + 9 = 0

and r1 + r 2 = 1 + 1 = 2. and x2 + y2 + 6x – 2y + 1 = 0
Thus, C1C2 > r1 + r2 Now, C1: (1, 3), C2: (–3, 1), r1 = 1 and r2 = 3.
Hence, the number of common tangents = 4. Therefore,

(iii) The given circles are C1C2 = (- 3 - 1)2 + (1 - 3)2
x2 + y2 – 6x – 6y + 9 = 0 = 16 + 4 = 20 = 2 5

andx2 + y2 + 6x + 6y + 9 = 0 and r1 + r2 = 1 + 3 = 4
Then, C1C2 > r1 + r2
Now, C1 = (3, 3), C2 = (–3, –3), r1 = 3 and r2 = 3. Thus, both the circles do not intersect.
Therefore, Hence, the number of common tangents = 4, in which
2 are direct common tangents and another 2 are trans-
C1C2 = (—3 - 2)2 + (-3 - 3)2 verse common tangents.
Here, the point M divides C2(–3, 1) and C1(1, 3) inter-
= 36 + 36 = 72 = 6 2 nally in the ratio 3 : 1.
Then, the co-ordinates of M are
and r1 + r2 = 3 + 3 = 6

Thus, C1C2 > r1 + r2

Hence, the number of common tangents = 4. Ê 3.1 - 1.3 , 3.3 + 1.1ˆ = Ê 0, 5ˆ
ËÁ 3 + 1 3 + 1 ¯˜ ÁË 2˜¯
(iv) The given circles are
Also, the point P divide C2(–3, 1) and C1 (1, 3) exter-
x2 + y2 – 4x – 4y = 0 nally in the ratio 3 : 1.
and x2 + y2 + 2x + 2y = 0

Now, C1 = (2, 2), C2 = (–1, –1), r1 = 2 2 and Then, the co-ordinates of P are

r2 = 2 Ê 3.1 + 1.3 , 3.3 - 1.1ˆ = (3, 4)
Therefore, ÁË 3 - 1 3 - 1 ˜¯

C1C2 = (-1 - 2)2 + (-1 - 2)2 Case I: Direct common tangents
= 9 + 9 = 18 = 3 2 Any line through (3, 4) is

and r1 + r2 = 2 2 + 2 = 3 2 y – 4 = m(x – 3) …(i)
Thus, C1C2 = r1 + r2 fi mx – y + 4 – 3m = 0
Hence, the number of common tangents = 3. If it is a tangent to the circle,
(v) The given circles are x2 + y2 = 64
and x2 + y2 – 4x – 4y + 4 = 0 m - 3 + 4 - 3m = 1
m2 + 1
Now, C1: (0, 0), C2: (2, 2), r1 = 8 and r2 = 2
Therefore, fi (1 – 2m)2 = (m2 + 1)
fi 3m2 – 4m = 0
C1C2 = (2 - 0)2 + (2 - 0)2 fi m(3m – 4) = 0
fi m = 0, 4 .
= 4+4=2 2
3
and r1 + r 2 = 8 + 2 = 10
Therefore, C1C2 < r1 + r2 Hence, the direct common tangents are
Thus, one circle lies inside of the other. y = 4, 4x – 3y = 0.
Hence, the number of common tangents = 0.
Case II: Transverse common tangents
Any line through (0, 5/2) is

3.46 Coordinate Geometry Booster

y - 5 = m(x - 0) …(i) fi –21 < – n2 < 39
2 fi –39 < n2 < 21

fi 2mx – 2y + 5 = 0 fi 0 < n < 21
If it is a tangent to a circle, then fi n = 1, 2, 3, 4

m◊1 - 3 + 5 Case II: When 5 < 3 + 25 - n2
2 =1
Then, 25 - n2 > 2
m2 + 1
fi 25 – n2 > 4
fi (2m – 1)2 = 4(m2 + 1) fi n2 < 21
fi 4m2 – 4m + 1 = 4(m2 + 1)
fi 4m + 3 = 0 fi n < 21

fi m ,m 3 fi n = 1, 2, 3, 4.
4
Hence, the number of integers = 4.
Hence, the equation of transverse tangents are x = 0 and
68. The given circles are
3x + 4y = 10.
66. Given circles are x2 + y2 + c2 = 2ax x2 + y2 + x + y = 0
and x2 + y2 + x – y = 0

fi (x - a)2 + y2 = ( a2 - c2 )2 …(i) Here, g1 = –1/2, f1 = –1/2, g2 = –1/2, f2 = 1/2, c1 = 0,
c2 = 0
and x2 + y2 + c2 = 2by Now, 2(g1 g2 + f1 f2) = 2(–1/4 + 1/4) = 0 and c1 + c2 = 0.
fi x2 + ( y — b)2 = ( b2 - c2 )2 Hence, the angle between the given circles is 90°.

…(ii) 69. The given circles are
x2 + y2 + 2a1x + 2b1y + c1 = 0
Now C1 : (a, 0), C2 : (0, b),
r1 = a2 - c2 and r2 = b2 - c2 and x2 + y2 + a2x + b2y + c2 = 0
Here, g1 = a1, f1 = b1, g2 = a2/2, f2 = b2/2, c1 = c1 and c2
Since both the circles touch each other externally, then = c2
C1C2 = r1 + r2 Since the given two circles are orthogonal, so

fi a2 + b2 = a2 - c2 + b2 - c2 2(g1g2 + f1f2) = c1 + c2

fi 2 Ê a1 ◊ a2 + b1 ◊ b2 ˆ = c1 + c2
ËÁ 2 2 ¯˜
fi a2 + b2 = a2 – c2 + b2 – c2 + 2( a2 - c2 )( b2 - c2 )
fi (a1 ◊ a2 + b1 ◊ b2) = c1 + c2
fi 2c2 = 2( a2 - c2 )( b2 - c2 ) Hence, the result.

fi c4 = (a2 – c2)(b2 – c2) = a2b2 – c2(a2 + b2) + c4 70. The given circles are
fi a2b2 – c2(a2 + b2) = 0
2x2 + 2y2 – 3x + 6y + k = 0

fi 1 + 1 = 1 fi x2 + y2 - 3 x + 3y + k = 0
a2 b2 c2 22

67. The given circles are and x2 + y2 – 4x + 10y + 16 = 0

x2 + y2 = 9 …(i) Since, the two given circles are orthogonal, then
and x2 + y2 – 8x – 6y + n2 = 0
2(g1g2 + f1f2) = c1 + c2

fi (x - 4)2 + ( y - 3)2 = ( 25 - n2 )2 …(ii) fi 2ÊÁË - 3 ¥ -2 + 3 ¥ 5ˆ˜¯ = k + 16
4 2 2
Here, C1: (0, 0), C2: (4, 3), r1 = 3 and r2 = 25 - n2
Since the circles have only two common tangents, so fi k + 16 = 18
we can write, 2

|r1 – r2| < C1C2 < r1 + r2 fi k = 18 — 16
2
fi |3 - 25 - n2 | < 5 < 3 + 25 - n2
fi k=4

Case I: When |3 - 25 - n2 | < 5 Hence, the value of k is 4.

71. Let the equation of the circle be

Then - 5 < 3 - 25 - n2 < 5 x2 + y2 + 2gx + 2fy = 0 …(i)

fi - 5 - 3 < - 25 - n2 < 5 — 3 Therefore, the centre of the circle is (–g, – f).

Since, the centre lies on the line y = x, so

fi — 2 < 25 - n2 < 8 –f = –g
fi 4 < 25 – n2 < 64
fi f=g

Circle 3.47

The Eq. (i) is orthogonal to 75. Hence, the equation of the radical axis is
x2 + y2 – 4x – 6y + 18 = 0,
(x2 + y2 + 4x + 6y + 9) –
so, 2[g(–2) + f(–3)] = 0 + 18 (x2 + y2 + 3x + 8y + 10) = 0
fi 2(–5g) = 18
fi g =-9 = f ( f = g) fi (4x – 3x) + (6y – 8y) + (9 – 10) = 0

5 fi x – 2y – 1 = 0

Hence, the equation of the circle is 76. Hence, the equation of the radical axis is
(x2 + y2 + 8x + 2y + 10) – (x2 + y2 + 7x + 3y + 10) = 0
x2 + y2 - 18 x - 18 y = 0
55 fi (8x – 7x) + (2y – 3y) = 0

fi y=x

72. Let the equation of the circle be Hence, the image of (2, 3) with respect to the line y = x
x2 + y2 + 2gx + 2fy + c = 0
…(i) is (3, 2). …(i)
77. Let S1 : x2 + y2 = 1 …(ii)
The equation of the circle (i) is orthogonal to …(iii)
S2 : x2 + y2 – 8x + 15 = 0
x2 + y2 + 4x – 6y + 9 = 0 and S3 : x2 + y2 + 10y + 24 = 0
and x2 + y2 – 5x + 4y – 2 = 0 Eq. (i) – Eq. (ii), we get

Thus, (4g – 6f) = c + 9 …(ii) 8 x = 16 fi x = 2

and (–5g + 4f) = c – 2 …(iii) Eq. (i) – Eq. (iii), we get

Subtracting, we get –9g + 10f + 11 = 0 –10 y = 25 fi y = –5/2

fi 9(–g) – 10(–f) + 11 = 0 Ê 5 ˆ
ËÁ 2 ¯˜
Hence, the locus of the centre is Hence, the radical centre is 2, - .

9x – 10y + 11 = 0

73. Let the equation of the circle be …(i) 78. Let the equation of the circle be …(i)
x2 + y2 + 2gx + 2fy + c = 0 x2 + y2 + 2gx + 2fy + c = 0 …(ii)

which is passing through (a, b). The circle (i) is orthogonal to

Therefore, x2 + y2 – 3x – 6y + 14 = 0,
x2 + y2 – x – 4y + 8 = 0,
a2 + b2 + 2ga + 2fb + c = 0 …(ii) and x2 + y2 + 2x – 6y + 9 = 0

It is given that the circle (i) is orthogonal to x2 + y2 = 4,

so 2(g ◊ 0 + f ◊ 0) = c – 4 fi c = 4 Therefore,

From Eq. (ii), we get 2 ÁËÊ g ◊ Ê - 3ˆ - 3f ˆ = c + 14
a2 + b2 + 2g ◊ a + 2f ◊ b + 4 = 0 ÁË 2¯˜ ¯˜

Hence, the locus of (–g, –f) is fi (–3g – 6f) = c + 14
2ax + 2by – (a2 + b2 + 4) = 0
È Ê 1ˆ ˘
74. Let C1 and C2 be the centres of the two orthogonal cir- 2 ÍÎ g ◊ ËÁ - 2˜¯ - 2f ˚˙ = c + 8

cles with radii r1 and r2, respectively. fi (2g – 4f) = c + 8 …(iii)
Here –C1PC2 = 90° and let
and
–PC1C2 = q and –PC2C1 = 90° – q
2(g ◊ 1 – 3 ◊ f) = c + 9
Thus,

sin (q ) = PM and sin (90° - q ) = PM . fi (2g – 6f) = c + 9 …(iv)
r1 r2
Eq. (iii) – Eq. (ii), we get

Squaring and adding, we get 5g + 2f = –6 …(v)

PM 2 PM 2 Eq. (iii) – Eq. (iv), we get
r1 r2
fi Ê ˆ + Ê ˆ = 1 2f = – 1 fi f = –1/2
ËÁ ˜¯ ËÁ ¯˜
Put the value of f in Eq. (v), we get,

PM 2 Ê 1 + 1ˆ =1 g = –1
ËÁ r12 r22 ¯˜
Also, put the values of f and g in Eq. (iv), we get

r12r22 c = –8.
r12 + r22
fi PM 2 = Hence, the equation of the circle is
x2 + y2 – 2x – y – 8 = 0

fi PM = r1r2 79. Equation of any circle passing through the point of in-
r12 + r22
tersection of the circles
Hence, the length of the common chord,
x2 + y2 + 13x – 3y = 0
PQ = 2PM = 2r1r2 and 2x2 + 2y2 + 4x – 7y – 25 = 0 …(i)
r12 + r22 is (x2 + y2 + 13x – 3y)

+ l(2x2 + 2y2 + 4x – 7y – 25) = 0

which is passing through (1, 1).

3.48 Coordinate Geometry Booster

Therefore, fi x2 + y2 + 2 Ê l - 3ˆ x
(1 + 1 + 13 – 3) + l(2 + 2 + 4 – 7 – 25) = 0 ÁË l + 1˜¯
fi 12 – 24l = 0
fi l=1 + 2 Ê 1 - 2l ˆ y + Ê 2(2 - 3l)ˆ = 0
ËÁ l +1 ˜¯ ËÁ l + 1 ˜¯
2
Put the value of l in Eq. (i), we get So its centre is Ê 3 —l , 2l — 1ˆ .
ËÁ l +1 l + 1 ˜¯
(x2 + y2 + 13x - 3y) + 1 (2x2 + 2y2 + 4x - 7y - 25) = 0
2 Since, the centre lies on the line y = x,

fi (2x2 + 2y2 + 26x – 6y) + (2x2 + 2y2 + 4x – 7y – 25) = 0 2l — 1 = 3 — l
fi (4x2 + 4y2 + 30x – 13y – 25) = 0 l +1 l +1

80. The equation of radical axis is fi 3l = 4
(x2 + y2 + 2x + 3y + 1)
–(x2 + y2 + 4x + 3y + 2) = 0 fi l = 4/3

fi (2x + 3y + 1) – (4x + 3y + 2) = 0 Hence, the equation of the circle is
fi 2x + 1 = 0 x2 + y2 - 10 x - 10 y - 12 = 0.
fi x=-1
777
2
83. The given circles are x2 + y2 + 2x + 3y + 1 = 0 and
Thus, the equation of the circle is x2 + y2 + 4x + 3y + 2 = 0.

(x2 + y2 + 2x + 3y + 1) + l(2x + 1) = 0 Hence, the equation of the common chord is 2x + 1 = 0.
fi (x2 + y2 + 2(l + 1)x + 3y + (1 + l)) = 0
Therefore, the equation of the circle is
Since AB is diameter of the circle, so the centre lies on (x2 + y2 + 2x + 3y + 1) + l(x2 + y2 + 4x + 3y + 2) = 0
it.
…(i)
Therefore, –2l – 2 + 1 = 0 fi (1 + l)x2 + (1 + l)y2 + 2(1 + 2l)x + 3(1 + l)y

+ (1 + 2l) = 0

fi l=-1 . fi x2 + y2 + 2 Ê 1 + 2l ˆ x + 3y + (1 + 2l) = 0
2 ÁË (1 + l)˜¯ (1 + l)

Hence, the equation of the circle is So, its centre is Ê - 1 + 2l , - 3ˆ
(x2 + y2 + 2x + 3y + 1) - 1 (2x + 1) = 0 ËÁ l +1 2¯˜
2
Since 2x + 1 = 0 is the diameter, so centre lies on it.
fi (x2 + y2 + 2x + 6y + 1) = 0
Therefore,

81. Any circle passing through the point of intersection of 2 ÊÁË - 1 + 2l ˆ +1= 0
the given line and the circle is l +1 ˜¯
x2 + y2 – 9 + l(x + y – 1) = 0
fi x2 + y2 + lx + ly – (9 + l) = 0 fi –2 – 4l + l + 1 = 0
fi l = -1
So the centre is Ê - l , - l ˆ .
ÁË 2 2 ¯˜ 3
Ê l l ˆ
Since, the circle is smallest, so the centre ËÁ - 2 , - 2 ˜¯ Hence, the equation of the circle (i) is
2(x2 + y2) + 2x + 6y + 1 = 0

lies on the chord x + y = 1.

Therefore, - l - l = 1 fi l = –1. LEVEL III
22
1. Given circle is x2 + y2 = 16.
Hence, the equation of the smallest circle is
x2 + y2 – 9 – (x + y – 1) = 0 Y

fi x2 + y2 – x – y – 8 = 0

82. The equation of any circle passing through the point of
intersection of the given circle is

(x2 + y2 – 6x + 2y + 4) + l(x2 + y2 + 2x – 4y – 6) = 0 X¢ O X

fi (1 + l)x2 + (1 + l)y2 + 2(l – 3)x + 2(1 – 2l)y
+ 2(2 – 3l) = 0

Y¢

Circle 3.49

Thus, the number of integral points inside the circle 4. Given circle is
x2 + y2 – 10x – 6y + 30 = 0
=5+7+7+7+7+7+5 S R

= 45 fi (x – 5)2 + (y – 3)2 C(5, 3)
= 25 + 9 – 30
2. Let S1: x2 + y2 + 2gx + 2fy + c = 0
and S2: x2 + y2 = 4 fi (x – 5)2 + (y – 3)2 = 4 P Q
S3: x2 + y2 – 6x – 8y + 10 = 0 Clearly, CP = radius = 2
S4: x2 + y2 + 2x – 4y – 2 = 0
Let PQ is parallel to y = x + 3

Now, S1 – S2 = 0 Therefore, the co-ordinates of P and R are obtained by

2gx + 2fy + c + 4 = 0 …(i) fi x -5 = y -3 = ±2
cos (45°) sin (45°)
Centre of a circle x2 + y2 = 4 lies on (i)

So, c + 4 = 0

c = –4 …(ii) x-5 y-3
1 1
Again, S1 – S3 = 0 fi = = ±2

2(g + 3)x + 2(f + 4)y + (c – 10) = 0 …(iii) 22

So, the centre of a circle x2 + y2 – 6x – 8y + 10 = 0 lies fi x-5= y-3=± 2
2
on (ii), so

fi 2(g + 3)3 + 2(f + 4)4 + (–4 – 10) = 0

fi 2(g + 3)3 + 2(f + 4)4 + –14 = 0 fi x-5= y-3=± 2

fi 6g + 8f + 18 + 32 – 14 = 0 fi x = 5 ± 2, y = 3 ± 2

fi 6g + 8f + 36 = 0

fi 3g + 4f + 18 = 0 …(iv) Thus, R = (5 + 2, 3 + 2)

Also, S1 – S4 = 0 and Q = (5 - 2, 3 - 2) .
fi 2(g – 1)x + 2(f + 2)y + (c + 2) = 0

So, centre (–1, 2) lies on (iii). 5. Clearly,

Thus, –2(g – 1) + 4(f + 2) + (–4 + 2) = 0 (2 2)2 + (2 2)2 = a2

fi –2(g – 1) + 4(f + 2) – 2 = 0 fi a2 = (2 2)2 + (2 2)2

fi (g – 1) – 2(f + 2) + 1 = 0 fi a2 = 8 + 8 = 16
fi |a| = 4
fi g – 2f = 4 …(v)
6. Given circle is
On solving, we get x2 + y2 – 8x + 2y + 12 = 0

g = –2 and f = –3 fi (x – 4)2 + (y + 1)2 = 16 + 1 – 12
fi (x – 4)2 + (y + 1)2 = 5
Hence, the equation of the circle is
As we know that the line is a chord, tangent or does not
x2 + y2 – 4x – 6y – 4 = 0
3. Given circle is (x ± a)2 + (y ± a)2 = a2 meet the circle at all, if

Y p < r, p = r or p > 4,

R(a, –a) Q(a, –a) where p is the length of the perpendicular from the cen-
X¢
tre to the line.
S(–a, –a) P(–a, –a)
So, p = 4+ 2-1 = 5 = 5 = r
1+ 4 5
X

Y¢ Thus, the line be a tangent to the circle.
7. It is given that
Hence, the radius of the smallest circle
= ( a2 + a2 - a), ( a2 + a2 + a) C1C2 = r1 + r2
= a 2 - a, a 2 + a
= ( 2 - 1)a, ( 2 + 1)a fi (a - b)2 + (b - a)2 = c + c

fi 2(a - b)2 = 2c
fi 2(a – b)2 = 4c2
fi (a – b)2 = 2c2

fi a -b = ±c 2

fi a=b±c 2

3.50 Coordinate Geometry Booster

8. Given circle is 10. The equation of the chord of contact of tangents to the
x2 + y2 + 4x – 6y – 12 = 0
(x + 2)2 + (y – 3)2 = 52 circle
x2 + y2 + 2gx + 2fy + c = 0
Thus, C = (–2, 3) and CA = 5 = CB
from (0, 0) is
and –CAB = p
3 x ◊ 0 + y ◊ 0 + g(x + 0) + f(y + 0) + c = 0

gx + fy + c = 0 …(i)

ËÊÁ p ˜¯ˆ MC Hence, the required distance from the point (g, f) to the
3 5
Now, sin = chord of contact (i)

C(–2, 3) Ê g2 + f 2 + cˆ
fi 3 = MC M(h, k) =Á ˜.
25 Ë g2 + f2 ¯

fi MC = 5 3 A B 11. We have,
2 (1 + ax)n = 1 + 8x + 24x2 + ◊◊◊

fi MC2 = 75 1 + n(a x) + n(n - 1) (a x)2 + ◊◊◊ = 1 + 8x + 24x2 + ◊◊◊
4 2

fi (h + 2)2 + (k - 3)2 = 75 Comparing the co-efficients, we get
4
fi na = 8, na (na - a ) = 24
Hence, the locus of M(h, k) is 2

(x + 2)2 + ( y - 3)2 = 75 8(8 - a ) = 24
4 2

9. fi (8 – a) = 6
fi a = 2, n = 4
Y Thus, the point P is (2, 4)

C(0, 3) Therefore, PA ◊ PB
OA = (PT)2 = 4 + 16 – 4 = 16

12. Given circles are x2 + y2 = 4
and x2 + y2 + 2x + 3y – 5 = 0

BX Hence, the common chord is

2x + 3y = 1 …(i)

Thus, the equation of the circle is

Let the equation of the circle be …(i) S + lL = 0
x2 + y2 + 2gx + 2fy + c = 0 fi (x2 + y2 – 4) + l(2x + 3l – 1) = 0
fi x2 + y2 + 2lx + 3ly – (l + 4) = 0
which touches the y-axis at C.
Put x = 0 in Eq. (i), we get So centre of the circle is

y2 + 2fy + c = (y – 3)2 Ê — l , — 3l ˆ
fi y2 + 2fy + c = y2 – 6y + 9 ËÁ 2 ¯˜
fi 2fy + c = –6y + 9
Comparing the co-efficients, we get Since the chord 2x + 3y = 1 is a diameter of the circle, so
- 2l - 9l = 1
2f = –6, c = 9
fi f = –3, c = 9 2
Intercepts on x-axis is
fi –13l = 2
fi 2 g2 - c = 8 fi l=- 2

fi g2 - c = 4 13

fi g2 - 9 = 4 Hence, the equation of the circle is

fi g2 – 9 = 16 x2 + y2 + 2 Ê - 2ˆ x + 3 Ê - 2ˆ y - Ê 4 - 2ˆ = 0
fi g2 = 25 ÁË 13¯˜ ËÁ 13¯˜ ËÁ 13¯˜
fi g = ±5
Hence, the equation of the circle is fi 13(x2 + y2) – 4x – 6y – 50 = 0

x2 + y2 ± 10x – 6y + 9 = 0 Therefore, the centre of the circle is Ê 2 , 3ˆ .
ÁË 13 13˜¯

Circle 3.51

13. Let (h, k) be a point lies on the circle and (x – 4)2 + (y – 4)2 = (32 – l)
x2 + y2 + 2gx + 2fy + c = 0 It is given that |r1 – r2| < C1C2 < r2 + r2

Thus, the length of the tangent from (h, k) | 2 - 32 - l | < 3 2 < ( 2 + 32 - l )

= h2 + k 2 + 2gh + 2fk + d Now, 3 2 < ( 2 + 32 - l )

= (h2 + k 2 + 2gh + 2fk) + d fi 32 - l > 2 2

= -c + d fi 32 – l > 8 …(i)
fi l < 24
= d -c and | 2 - 32 - l | < 3 2

14. Equation of the circle is fi - 4 2 < - 32 - l < 2 2

S + lL = 0 fi —2 2 < 32 - l < 4 2

fi (x2 + y2 – 4) + l(x + y – 1) = 0 fi 8 < (32 – l) < 32

fi x2 + y2 + lx + ly – (l + 4) = 0 fi –24 < –l < 0

So centre of the circle is Ê - l ,- lˆ fi 0 < l < 24 …(ii)
ÁË 2 2 ˜¯
From Eqs (i) and (ii), we get

Since the chord x + y = 1 is a diameter of the circle, so 0 < l < 24
-l - l =1
18. Since two vertices of an equilateral triangle are
22
B(–1, 0) and C(1, 0).

So, the third vertex must lie on the y-axis.

fi –l = 1 Let the third vertex be A(0, b).
fi l = –1
Now, AB = BC = CA
Hence, the equation of the circle is fi AB2 = BC2 = AC2
x2 + y2 – x – y – 3 = 0 fi 1 + b2 = 4 = 1 + b2
fi b2 = 4 – 1
15. Now the image of the centre (3, 2) to the line x + y = 19
fi b= 3
is obtained by

a - 3 = b - 2 = - 2 Ê 3 + 2 - 19ˆ Thus, the third vertex is A = (0, 3)
1 1 ËÁ 12 + 12 ¯˜ As we know that in case of an equilateral triangle,

a - 3 = b - 2 = 14 Circumcentre = Centroid = Ê 0, 1ˆ
11 ËÁ 3 ˜¯

a = 17, b = 16 Hence, the equation of the circumcircle is

Hence, the equation of the new circle is Ê 1 ˆ 2 Ê 1 ˆ2
(x – 17)2 + (y – 16)2 = 1 ÁË 3 ¯˜ ËÁ 3 ¯˜
fi (x — 0)2 + y - = (1 - 0)2 + 0 -

16. Given circles are Ê 1 ˆ 2 4
(x – 1)2 + (y – 3)2 = r2 ÁË 3 ˜¯ 3
…(i) fi x2 + y - =
and x2 + y2 – 8x + 2y + 8 = 0 …(ii)
fi (x – 4)2 + (y + 1)2 = 16 + 1 – 8 19. The vertices of the triangle are
fi (x – 4)2 + (y + 1)2 = 9
A = (0, 6), B = (2 3, 0), C = (0, 2 3)

Two circles (i) and (ii) intersect, if Let P(h, k) be the circumcentre, then
PA = PB = PC
|r1 – r2| < C1C2 < r1 + r2
fi PA2 = PB2 = PC2
fi |r — 3| < (1 - 4)2 + (3 + 1)2 < r + 3
fi h2 + (k - 6)2 = (h - 2 3)2 + k 2
fi |r – 3| < 5 < r + 3
fi |r – 3| < 5 and r + 3 > 5 = h2 + (k - 2 3)2
fi –5 < (r – 3) < 5, r > 2
fi –2 < r < 5 + 3, r > 2 On solving, we get
fi r < 8, r > 2
fi 2<r<8 h = 0 and k = 2
17. Since there are two real tangents drawn, so the circles
intersect. Thus, radius, r = h2 + (k - 6)2 = 4
Given circles are (x – 1)2 + (y – 1)2 = 2
Hence, the equation of the circle is
x2 + (y – 2)2 = 16

fi x2 + y2 – 4y – 12 = 0

3.52 Coordinate Geometry Booster

20. Given circle is 24. Let ABC be an equilateral triangle with AB = a.

x2 + y2 + 4x – 6y + 4 = 0 P
(x + 2)2 + (y – 3)2 = 32

So, C = (–2, 3) and r = 3 C(–2, 3) DC
As we know that the centroid

divides the median in the ratio O

2:1

Here, circumradius = 2 and in- AB
QM
radius = 6 R

Hence, the equation of the circumcircle is Let PM = p
(x + 2)2 + (y – 3)2 = 62
Now, sin (60°) = PM = p
fi x2 + y2 + 4x – 6y – 23 = 0 PQ a

21. Given circle is fi p= 3
a2
x2 + y2 + 2gx + 2fy + c = 0 D C
a
We have, B p= 3a
2
a2 + a2 = (r + r)2 O B fi

fi a2 = 2r2 r Here O is the centroid.

fi a2 = 2r2 So the centroid divides the median in the ratio 2 : 1
fi a=r 2
Aa Thus, OM = p
fi r= p 3
Hence, the result.
3
22. Given circle is
x2 + y2 + 2gx + 2fy + c = 0 fi 2r = 2p = 2 ¥ 3 a = a
332 3
Here, OA = r, –AOM = 60°
Let x be the side of a square.
C
Now, O Thus,
sin (60°) = AM = AM
OA r M Ê aˆ 2 a2
ÁË 3 ˜¯ 3
x2 + x2 = =

fi AM = 3 A fi 2x2 = a2
r2 3

fi AM = 3 r fi x2 = a2
2 6 a2

fi AB = 2AM = r 3 Area of the square =
6
Hence, the area of an equilateral triangle is
25. Given circle is
= 3 ¥ ( AB)2 x2 + y2 – 12x + 4y + 30 = 0
4
fi (x – 6)2 + (y + 2)2 = 36 + 4 – 30
= 3 ¥ 3r2 fi (x - 6)2 + ( y + 2)2 = ( 10)2
4
Any point on the circle is
= 3 3 r2
4 P(6 + 10 cos q, - 2 + 10 sin q )

23. Given centre of the circle is (0, 0). Let d be the distance from the origin
As we know that the centroid divides the median in the Thus, d = OP
ratio 2 : 1
\ Circumradius = 2a d 2 = (6 + 10 cos q )2 + (- 2 + 10 sin q )2
Hence, the equation of the circle is
x2 + y2 = (2a)2 = 50 + 4 10(3 cos q - sin q )
fi x2 + y2 = 4a2
= 40 + 4.10ËÁÊ 3 cos q - 1 sin q ˜ˆ¯
10 10

= 40 + 40(cos q cos a – sin q sin a)

= 40 + 40(cos(q + a))

= 40 + 40, when cos(q + a) = 1 = cos 0°

= 80 and q = –a

Circle 3.53

Hence, the point is fi mx – y – m = 0 …(ii)
= (6 + 10 cos q, - 2 + 10 sin q ) If (ii) is a tangent of (i), then
fi m-2-m = 5
= Ê 6 + 10 ◊ 3 ,-2+ 10 ◊ Ê - 1 ˆˆ
ÁË 10 ÁË 10 ˜¯ ˜¯ m2 + 1

= (6 + 3, –2 – 1) fi -2 = 5
= (9, –3) m2 + 1
26.
27. The equation of any tangent through origin is fi 5(m2 + 1) = 2
y = mx fi (m2 + 1) = 2
If y = mx be a tangent to the given circle, then
5
7m + 1 = 5 fi m2 = 2 - 1 = - 3
m2 + 1
55
fi m=j

fi 25(m2 + 1) = (7m + 1)2 So, no real tangents can be drawn.
fi 25m2 + 25 = 46m2 + 14m + 1
fi 24m2 – 14m – 24 = 0 Thus, number of tangents = 0.
30. The equation of the tangent to the circle x2 + y2 = a2

Let its roots are m1, m2. at (a cosq, a sinq ) and ÎÈÍa cos Ê p + q ˆ , a sin Ê p + q ˆ ˘
So, product of the roots = –1 ÁË 3 ¯˜ ÁË 3 ˜¯ ˙˚
fi m1 ◊ m2 = –1
fi q=p are x cos q + y sin q = a

2 and x cos ÊÁË p + q ˆ¯˜ + y sin ËÁÊ p + q ˜¯ˆ = a
3 3

Let (h, k) be the point of intersection.

28. Given circle is On solving, we get

x2 + y2 – 2x – 4y = 0 2a Ê sin Ê p + q ˆ - sin q ˆ
fi (x – 1)2 + (y – 2)2 = 5 ÁË ÁË 3 ˜¯ ˜¯
…(i) h =

Equation of any tangent passing through (4, 3) is 3

y – 3 = m(x – 4) 2a Ê cos Ê p + q ˆ - cos q ˆ
ÁË ËÁ 3 ¯˜ ¯˜
fi mx – y + (3 – 4m) = 0 …(ii) and k =

If (ii) is a tangent of (i), then 3

fi m - 2 + 3 - 4m = 5 Now, squaring and adding, we get
m2 + 1
3h2 + 3k 2 =1
4a2 4a2
1 — 3m = 5
m2 + 1 fi h2 + k 2 = 4a2
3

fi 5(m2 + 1) = (1 – 3m)2 Hence, the locus of (h, k) is
fi 5m2 + 5 = 1 – 6m + 9m2
fi 4m2 – 6m – 4 = 0 x2 + y2 = 4a2
fi 2m2 – 3m – 2 = 0 3

Let its roots are m1, m2. 31. The equation of tangent to the circle x2 + y2 = 4 at
So, product of the roots = –1
(1, 3) is x + 3y = 4
fi m1 ◊ m2 = –1
Y
fi q=p
2 B
P(1, 3)

29. Given circle is …(i) OM AX

x2 + y2 – 2x – 4y = 0
fi (x – 1)2 + (y – 2)2 = 5

The equation of any tangent passing through (0, 1) is
y = m(x – 1)

3.54 Coordinate Geometry Booster

Clearly, A = (4, 0), B = (0, 4/ 3) (ii) Area of the sector OAC

Thus, the area of the triangle OPA is = 1 r2q
2
= 1 ¥4¥ 3 = 2 3 sq. u.
2 1 Ê 13ˆ 2 Ê 120ˆ
= 2 ÁË 2 ¯˜ sin -1 ÁË 169¯˜

LEVEL IV = 169 sin -1 Ê 120ˆ sq. m.
8 ÁË 169˜¯
1. (i) Here, AB = 13 m and AC = 5 m
Area of the triangle OAC = 1 r2 sin q
Y 2

C 1 Ê 13ˆ 2 Ê 120ˆ
2 ËÁ 2 ˜¯ ÁË 169˜¯
q =
OA
X¢ B X = 15 sq. m.

C¢ Thus, the area of the smaller portion bounded by the
circle and the chord AC.

= Area of sector OAC – Area of DAOC

Y¢ = Ï169 sin -1 Ê 120 ˆ - 15¸˝ sq. m.
Ì ËÁ 169 ˜¯ ˛
Ó 8
13 13
Let the co-ordinates of C be Ê 2 cos q, 2 sin q ˆ 2. Here, the co-ordinates of the centres A, B and C are
It is given that AC = 5 ÁË ¯˜
(0, 0), ( 55, 3) and ( 65, 4).

fi AC2 = 25 Y

Ê 13 13ˆ 2 Ê 13 ˆ 2
ÁË 2 2 ˜¯ ÁË 2 ˜¯
fi cos q - + sin q = 25

fi cos q = 238 = 119 B
338 159
X¢ A D X
E

fi sin q = 1- Ê 119ˆ 2 = 120 C
ÁË 159˜¯ 169
Y¢
fi q = sin -1 Ê 120ˆ
ËÁ 169¯˜ 3. The equation of the radical axis is
(x2 + y2 + 4x + 2y + 1)

The co-ordinates of C and C¢ will become - Ê x2 + y2 — x + 3y — 3ˆ = 0
ËÁ 2˜¯
Ê 119 , 60ˆ and Ê 119 , - 60 ˆ
ÁË 26 13 ¯˜ ËÁ 26 13 ¯˜ fi 5x - y + 5 = 0
2
Thus, the co-ordinates of B are Ê - 13 , 0ˆ˜¯
ÁË 2 10x – 2y + 5 = 0 …(i)

Hence, the equations of the pair of lines BC and BC¢ The equation of the co-axial circle is
are
ÊËÁ x2 + y2 - x + 3y - 32ˆ¯˜ +
± (60/13) Ê 13ˆ
y — 0 = ÁË x + 2 ¯˜ l(x2 + y2 + 4x + 2y + 10) = 0
Ê 119 13ˆ
ÁË 26 + 2 ¯˜ Ê 4l -1ˆ Ê 2l + 3ˆ Ê l - 3/2ˆ
ËÁ l +1 ¯˜ ËÁ l +1 ¯˜ ËÁ l +1 ˜¯
15 ÁËÊ 123ˆ˜¯ x2 + y2 + x + y + = 0
36
fi y = ± x + …(ii)

Thus, È1 Ê 4l - 1ˆ , 1 Ê 2l + 3ˆ ˘
±5 Ê 13ˆ Í ÁË l + 1 ¯˜ ˙
fi y = 12 ËÁ x + 2 ¯˜ Î 2 2 ÁË l + 1 ¯˜ ˚

fi 24y = ±5(2x + 13) Clearly the centre lies on the radical axis.

Circle 3.55

So Ê 1 Ê 4l - 1ˆ ˆ + 2 Ê 1 Ê 2l + 3ˆ ˆ + 5 = 0 y - 0 = 7(x - 2) and y - 0 = 1 (x - 2)
10 ÁË 2 ËÁ l + 1 ˜¯ ¯˜ ËÁ 2 ËÁ l + 1 ¯˜ ˜¯ 7

fi l = 1. fi 7x + y = 14 and x – 7y = 2
Put the value of l = 1 in Eq. (ii), we get
Now, the centres of the circles lie on the lines AB and
4x2 + 4y2 + 6x + 10y – 1 = 0. AC at a distance of 5 2 units.

4. A circle passing through the point of intersection of Thus C1: x-2 = y - 0 = 5 2
x2 + y2 + ax + by and Ax + By + C = 0 is 1 7
x2 + y2 + ax + by + c + l(Ax + By + C) = 0 …(i)
52 52
and a circle passing through the point of intersection
x2 + y2 + a¢x + b¢y + c¢ = 0 and (x, y) = (1, 7)
A¢x + B¢y + C¢ = 0 is
x2 + y2 + a¢x + b¢y + c¢ + l¢ (a¢x + B¢y + C¢) = 0 and C2: x-2 = y - 0 = 5 2
7 1
…(ii)
52 52
Since the point of intersection lie on the circle, so,
(x, y) = (9, 1)
a + lA = a¢ + l¢A¢
b + lB = b¢ + l¢B¢ Hence, the equations of the circles are
and c + lC = c¢ + l¢C¢
Eliminating l and l¢, we get (x – 1)2 + (y – 7)2 = 9
or (x – 7)2 + (y – 1)2 = 9
A A¢ a — a¢
6. Let the centre of the circle be (h, k) such that k = h – 1.
fi B B¢ b — b¢ = 0 We have, (h – 7)2 + (k – 3)2 = 32
fi (h – 7)2 + (h – 4)2 = 32
C C¢ c — c¢ fi h2 = 11h + 28 = 0

B¢ b — b¢ A¢ a — a¢ A¢ a — a¢ fi h = 7, 4 and k = 6, 3
fi A -B +C =0
Hence, the equation of a circle can be
C¢ c — c¢ C¢ c — c¢ B¢ b —b¢ (x – 7)2 + (y – 6)2 = 32

fi AB¢(c - c¢) - AC¢(b - b¢) - BA¢(c - c¢) and (x – 4)2 + (y – 3)2 = 32
+ BC¢(a - a¢) + CA¢(b - b¢) - CB¢(a - a¢) = 0 fi x2 + y2 – 8x – 6y + 16 = 0
and x2 + y2 – 8x – 6y + 16 = 0
fi (a - a¢)(BC ¢ - CB¢) + (b - b¢)(CA¢ - AC ¢)
+ (c - c¢)( AB¢ - BA¢) = 0 7. Here, CM = CN

5. The equation of the tangent to the given circle Y
(x + 2)2 + (y + 3)2 = 25 at A(2, 0) is
4x – 3y – 8 = 0 …(i) P(2, 8) N

P X¢ C(h, k)
O
B M

X

A(2, 0)

TC Y¢

Let m and m¢ be the slopes of the lines AB and AC. fi 4h - 3k - 24 = 4h + 3k - 42
Thus, tan (45°) = m - (4/3) 16 + 9 16 + 9

1 + m ◊ (4/3) fi (4h – 3k – 24) = ±(4h + 3k – 42)
and tan (135°) = m¢ - (4/3)
fi k=3
1 + m¢ ◊ (4/3)
On solving, we get Also, r = CM

m = - 7, m¢ = 1 fi r2 = CM 2
7
ÁËÊ 4h + 3k - 42˜ˆ¯ 2
Therefore, the equations of the lines AB and AC are 5
fi (h — 2)2 + (k - 8)2 =

3.56 Coordinate Geometry Booster

ËÊÁ 4h + 9 - 42˜ˆ¯ 2 The equation of the normal is
5
fi (h — 2)2 + (3 - 8)2 = ( y - 4) = - 1 (x - 2)
4
ËÁÊ 4h — 33˜ˆ¯ 2
5 fi 4(y – 4) = –(x – 2)
fi (h — 2)2 + 25 = fi x + 4y = 18
Let (h, k) be the centre of the circle.
On solving we get, Thus, h + 4 k = 18
Also, CP = CQ
h=2 fi h2 + (k – 1)2 = (h – 2)2 + (k – 4)2 …(i)
fi –2k + 1 = –4h + 4 – 8k + 16 …(ii)
Thus, the equation of the circle is fi 4h + 6k = 19
On solving Eqs (i) and (ii), we get
(x – 2)2 + (y – 3)2 = 25
fi x2 + y2 – 2x – 6y – 12 = 0 h = - 16 , k = 53 16 53ˆ
8. Let x1, x2 are the roots of x2 + 2x – a2 = 0. 5 10 5 10 ¯˜
Then x1 + x2 = –2, x1x2 = –a2
Similarly, y1 + y2 = –4, y1y2 = –b2 Hence, the required centre is Ê , .
ËÁ
Hence, the equation of the circle is

(x – x1)(x – x2) + (y – y1)(y – y2) = 0 11. Let the mid point be M(h, k).
x2 – (x1 + x2)x + x1x2 + y2 – (y1 + y2)y + y1y2 = 0
x2 + 2x – a2 + y2 + 4y – b2 = 0 Equation of the chord bisected at M is

x2 + y2 + 2x + 4y – (a2 + b2) = 0 T = S1
hx + ky – (x + h) – 3(y + k) = h2 + k2 – 2h – 6k
Therefore, the centre of the circle = (–1, –2) and the

radius = a2 + b2 + 5 . which is passing through the origin. So,
9. Equation of the tangent to the circle x2 + y2 = 5 at (1, –h – 3k = h2 + k2 – 2h – 6k

–2) is x – 2y = 5. fi h2 + k2 – h – 3k = 0

Given circle is Hence, the locus of (h, k) is
x2 + y2 – 8x + 6y + 20 = 0 x2 + y2 – x – 3y = 0

Centre = (4, –3) and radius = 5 . …(i) 12. The equation of a circle passing through the points of
…(ii) intersection of x2 + y2 = 1
Let the point of contact be (h, k). x2 + y2 – 2x – 4y + 1 = 0 is

Thus, h – 2 k = 5 (x2 + y2 – 2x – 4y + 1) + l(x2 + y2 – 1) = 0
and (h – 4)2 + (k + 3)2 = 5
fi h2 + k2 – 8h + 6k = 20 = 0 fi (1 + l)x2 + (1 + l)y2 – 2x – 4y + (1 – l) = 0

On solving Eqs (i) and (ii), we get fi x2 + y2 - 2 x - 4 y + (1- l) = 0 …(i)
(1+ l) (1+ l) (1+ l)
h = 3, k = –1
Centre = Ê (1 1 l) , 2ˆ
Hence, the point of contact is (3, –1). ËÁ + (1 + l)˜¯
10. Given curve is y = x2

Y Ê 1 ˆ 2 Ê 2 ˆ 2 Ê (1 - l ) ˆ 2
ËÁ + ¯˜ ËÁ + ˜¯ ÁË (1 + l ) ¯˜
NT Radius = (1 l ) + (1 l ) +

C(h, k) Equation (i) touching the straight line x + 2y = 0. So,
Q(0, 1) P(2, 4) the length of perpendicular from the centre is equal to
the radius of a circle.

X¢ O X Thus, 5 = Ê 1 ˆ 2 Ê 2 ˆ 2 Ê (1 — l ) ˆ 2
(1 + l) ËÁ + ˜¯ ËÁ + ˜¯ ËÁ (1 + l ) ¯˜
(1 l ) + (1 l ) +

fi dy = 2x fi (1 – l)2 + 5 = 5
dx fi l=1

Thus, m = Ê dy ˆ = 4 Hence, the equation of the circle is
ÁË dx ¯˜ /(2,4) x2 + y2 – x – 2y = 0

13. As we know that the normal always passes through the

centre of the circle.

So, the equation of the circle is
(x – 3)2 + y2 = 6

Circle 3.57

14 Let the equations of the two given circles are Also, (h – 1)2 + (k – 2)2 = (h – 5)2 + (k – 2)
fi (h – 1)2 = (h – 5)2
x2 + y2 + 2gx + g2 = 0 fi h2 – 2h + 1 = h2 – 10h + 25
and x2 + y2 + 2fy + f 2 = 0 fi 8h = 24
fi h = 3.
whose centres lie on x and y axes. Thus, the centre is (3, 0)
Hence, the radius of the circle is 2 2.
Thus, the equation of the radical axis is
18. Let the co-ordinates of P be (0, k) and the centre of the
2gx + g2 – 2fy – f 2 = 0 circle be C(h, k).
fi 2gx – 2fy + g2 – f 2 = 0
Y
15. Equations of the normals of the circle are
A
x + 2xy + 3x + 6y = 0

fi (x + 2y)(x + 3) = 0

fi (x + 2y) = 0, (x + 3) = 0 Ê 3ˆ
ÁË 2¯˜
Thus, the centre of the circle is C1: - 3, .
Given circle is

x(x – 4) + y(y – 3) = 0 P B
fi x2 + y2 – 4x – 3y = 0 X¢ O X

Centre is C2: Ê 2, 3ˆ and the radius is 5.
ËÁ 2˜¯ 2

According to the questions, we get Y¢

C1C2 = r1 – r2 Now, CA = CB = CP
fi CA2 = CB2 = CP2
fi (- 3 - 2)2 + 0 = r — 5 fi (h – 4)2 + (k – 3)2 = (h – 2)2 + (k – 5)2 = h2
2
Now,
fi r = 5 + 5 = 15
22 (h – 4)2 + (k – 3)2 = (h – 2)2 + (k – 5)2

Hence, the equation of the circle is fi –8h – 6k = –4h + 4 – 10k

ÁÊË 23˜ˆ¯ 2 Ê 15ˆ 2 fi 4h – 4k + 4 = 0
ÁË 2 ¯˜
(x + 3)2 + y - = fi h–k+1=0 …(i)
and (h – 2)2 + (k – 1)2 = h2
fi x2 + y2 + 6x – 3y – 45 = 0 fi (h – 2)2 + (h – 4)2 = h2

16. Here A = (2, 3), B = (6, 3) and D = (2, 6) fi h2 – 4h + 4 + h2 – 8h + 16 = h2
fi –4h + 4 + h2 – 8h + 16 = 0
Thus, the centre of the circle is (4, 9/2) and the radi- fi h2 – 12h + 20 = 0

us = 5/2.

Y fi (h – 2) = 0, (h – 10) = 0

D(2, 6) fi (h – 2) = 0, (h – 10) = 0

fi h = 2, 10

C(4, 9/2) Thus, k = h + 1 = 3, 11

Hence, the point on y-axis is P (0, 3).

y=3 Thus, the equation of the circle is
B(6, 3)
A(2, 3) (x – 2)2 + (y – 3)2 = 9
O
X fi x2 – 4x + 4 + y2 – 6y + 9 = 9
x=2 fi x2 + y2 – 4x – 6y + 4 = 0

19. Let the centre of the circle be C(h, k).

Hence, the equation of the circle is Y

Ê 9ˆ 2 25 P(3, 3)
ÁË 2˜¯ 4
(x - 4)2 + y - =

y=x

fi x2 + y2 – 8x – 9x + 30 = 0 O 3x – y = 6 X
X¢ B
17. Let the centre be (h, k)
Thus, (h – 1)2 + (k – 2)2 = (h – 5)2 + (k – 2)2 A(1, –3)
= (h – 5)2 + (k + 2)2
Now, (h – 5)2 + (k – 2)2 = (h – 5)2 + (k + 2)2 Y¢
fi (k – 2)2 = (k + 2)2
fi k2 – 4k + 4 = k2 + 4k + 4 Here, PA = 2 10
Let B be (a, a).
fi 8k = 0

fi k=0

3.58 Coordinate Geometry Booster

Thus, PB = 2 10 fi x2 - 9x + 81 + y2 - 2ky = 25
fi (a – 3)2 + (a – 3)2 = 40 44
fi (a – 3)2 = 20
fi (a - 3) = ± 2 5 fi x2 + y2 – 9x – 2ky + 14 = 0
fi a=3±2 5 23. Let P be (h, k)
fi a=3-2 5
Hence, B = (3 - 2 5, 3 - 2 5) . CB
20. The length of the line = the length of the tangent B(h, k)

= 4 + 9 + 4 + 30 + 1 A Q(6, –1)
x+y=5
= 48 = 4 3
21. Let the centre be C(h, h)

Y

M 4x + 3y = 6 Here, PA = PB = PC
h fi PA2 = PB2 = PC2
h C(h, h) fi h2 + k2 + h – 3 = 3(h2 + k2) – 5h + 3k

X¢ O X = 4(h2 + k2) + 8h + 7k + 9
On solving, we get
Y¢
h = 0, k = –3.
Now, CM = h So, the point P is (0, –3).
Let the centre of the new circle be (a, b).
fi 4h + 3h = 6
5
We have,
fi 7h = 30
fi h = 30 a + b - 5 = (a - 6)2 + (b + 1)2
2
7 = a2 + (b + 3)2

Hence, the equation of the circle is

Ê 30ˆ 2 Ê 30ˆ 2 Ê 30ˆ 2 On solving, we get
ÁË 7 ˜¯ ÁË 7 ¯˜ ËÁ 7 ¯˜ a= 7,b= -7
fi x - + y - = 22

fi x2 - 60 x + 900 + y2 - 60 y = 0
7 49 7
and radius is r = 5 2
fi x2 + y2 - 60 x - 60 y + 900 = 0 2
7 7 49
Hence, the equation of the circle is
fi 49(x2 + y2) – 420(x + y) + 900 = 0
2
Ê 7ˆ 2 Ê 7ˆ 2 Ê 5 2 ˆ
ËÁ 2˜¯ ÁË 2¯˜ ÁË 2 ˜¯
22. Hence, the equation of the circle is x - + y + =

ÊËÁ 9 ˜¯ˆ 2 Ê 9 2˜¯ˆ 2
2 ÁË 2
x - + ( y - k)2 = - + k2 fi x2 + y2 – 7x + 7y + 12 = 0

Y 24. Clearly, the locus of the mutually perpendicular tan-
gents to the circle is the director circle. So its equation
C(9/2, k) is x2 + y2 = 18.
Thus, its centre is (0, 0) and the radius = 3 2 .
O A(2, 0) B(7, 0) X Let the equation of the co-axial system be

x2 + y2 + 2gx + c = 0

whose centre is (–g, 0) and the radius = g 2 - c .

Circle 3.59

P( 2, 4) C¢(–g, 0) fi 4(y – 4) = –(x – 2) …(i)
C(0, 0) fi x + 4y = 18 …(ii)
Let (h, k) be the centre of the circle.
Clearly, — g + 0 = 2 Thus, h + 4 k = 18
2 Also, CP = CQ
fi h2 + (k – 1)2 = (h – 2)2 + (k – 4)2
fi g = -2 2 fi –2k + 1 = –4h + 4 – 8k + 16
Also, CC¢ = r1 + r2 fi 4h + 6k = 19
fi g = 3 2 + g2 - c On solving (i) and (ii), we get

h = - 16 , k = 53
5 10

Hence, the required centre is Ê - 16 , 53ˆ .
ËÁ 5 10 ¯˜

fi —2 2 = 3 2 + 8 - c 27. Given equation of normals are
x2 – 3xy – 3x + 9y = 0
fi —5 2 = 8 - c

fi 8 – c = 50 fi (x – 3y)(x – 3) = 0
fi c = –42
Hence, the equation of the co-axial circle is fi (x – 3y) = 0, (x – 3) = 0
x2 + y2 - 4 2x - 42 = 0 .
25. Equation of the tangent to the circle x2 + y2 = 5 at As we know that the normal always passes through the
(1, –2) is x – 2y = 5. centre of a circle.
Given circle is
Thus, centre, C1 = (3, 1)
x2 + y2 – 8x + 6y + 20 = 0 Also, given circle is

Centre = (4, –3) and the radius = 5. x2 + y2 – 6x + 6y + 17 = 0
fi (x – 3)2 + (y + 3)2 = 1
Let the point of contact be (h, k). …(i)
Thus, h – 2 k = 5 …(ii) So the centre C2 = (3, –3) and r2 = 1
and (h – 4)2 + (k + 3)2 = 5 Since the circle touches externally, so
fi h2 + k2 – 8h + 6k + 20 = 0
C1C2 = r1 + r2
On solving Eqs (i) and (ii), we get fi 4 = r1 + 1
fi r1 = 3
h = 3, k = – 1 Hence, the equation of the circle is

Hence, the point of contact is (3, –1). (x – 3)2 + (y – 1)2 = 9
fi x2 + y2 – 6x – 2y + 1 = 0
26. Given curve is
28. Given lines are 2x – 4y = 9
Y

NT and 2x - 4y + 7 = 0
3
C(h, k) 1 9+ 7
Q(0, 1) P(2, 4) 3
Hence, the radius = 2
4 + 16
X¢ O X

= 1 Ê 3 34 ˆ
2 ÁË 20 ¯˜

y = x2 = 1 Ê 34 ˆ = 17
2 ÁË 6 5 ¯˜ 65
fi dy = 2x
dx 29. Let the co-ordinates of the other end be (a, b).
Given that the centre of the given circle = (4, 2).
Ê dy ˆ
Thus, m = ÁË dx ˜¯ /(2,4) = 4 Thus, a - 3 = 4, b + 2 = 2
22
The equation of the normal is
a = 11, b = 2
( y - 4) = - 1 (x - 2) Hence, the other end be (11, 2).
4

3.60 Coordinate Geometry Booster
30.
Hence, the equation of the line is

P Ê 2 - 1ˆ
y = ÁË 2 + 1˜¯
x = (3 — 2 2)x

DC 32. Given circle is x2 + y2 – 6x – 4y + 4 = 0

O Y

AB R P(h, k)
QM B

Let PM = p C X
Now, sin (60°) = PM = p A

PQ a O

fi p= 3
a2

fi p= 3a So, the centre is (3, 2) and the radius is 3.
2 Let P be (h, k)
Thus, 4h – 3k = 6
Here O is the centroid. Let –CPB = q …(i)
Then sin q = BC …(ii)
So the centroid divides the median in the ratio 2 : 1. …(iii)
PC
Thus, OM = p =3
3
(h - 3)2 + (k - 2)2
fi r= p
3

fi 2r = 2p = 2 ¥ 3 a = a From Eqs (i) and (ii), we get
332 3

Let x be the side of a square. sin q = 3

Ê aˆ 2 a2 Ê 4h — 6 2ˆ˜¯ 2
ÁË 3 ˜¯ 3 ÁË 3
Thus, x2 + x2 = = (h - 3)2 + -

fi 2x2 = a2 It is given that,
3
-1 Ê 24ˆ
fi x2 = a2 2q = tan ËÁ 7 ˜¯
6 a2
fi tan (2q ) = 24
Area of a square = 6 7

31. Given circle is x2 + y2 = 3 fi cos (2q ) = 7
and the line is x + y = 2 25

Now, fi 1 — 2 sin2 (q ) = 7
x2 + (2 – x)2 = 3 25

fi x2 + x2 – 4x + 4 = 3 fi 2 sin2 (q ) = 1 — 7 = 18
25 25
fi 2x2 – 4x + 1 = 0
fi sin2 (q ) = 9
fi x = 4± 8 =1± 1 25
42
fi sin (q ) = 3
fi x =1+ 1 5
2
From Eq. (iii), we get
Also, y = 2 — x = 2 - 1 - 1 = 1 - 1
22 3= 3
5 2
ËÊÁ 4h — 6 2ˆ˜¯
Thus, the point of intersection is (h - 3)2 + 3 -

ÁËÊ1 + 1 ,1- 1ˆ Ê 4h — 6 2ˆ¯˜ 2
2 2 ¯˜ ËÁ 3
fi (h - 3)2 + - = 25

Circle 3.61

ÊÁË 4h — 12 ¯ˆ˜ 2 fi cos ËÁÊq — p ˜¯ˆ = cos Ê 3p ˆ
3 4 ËÁ 4 ¯˜
fi (h - 3)2 + = 25

fi 9(h – 3)2 + (4h – 12)2 = 225 fi q = p or — p
fi 25h2 – 150h = 0 2
fi h = 0, 6
If h = 0, then k = –2. Hence, the point B be (–2, 2) or (2, –2).
If h = 6, then k = 6. 35. Given circle is x2 + y2 – 2rx – 2hy + h2 = 0
Hence, the points are (0, –2), (6, 6).
fi (x – r)2 + (y – h)2 = r2.
33. Here, C1 = (5, 0), r1 = 3, C2 = (0, 0), r2 = r
It is given that two circles intersect. So, the centre is (r, h) and radius is r.

It is possible only when r = h.

So, |r1 – r2| < C1C2 < r1 + r2 C(r, h) C(r, r)
fi |3 – r| < 5 < 3 + r r
fi |3 – r| < 5, 5 < 3 + r
fi –5 < (r – 3) < 5, r > 2
fi –2 < r < 8, r > 2
fi 2<r<8
34. Given circle is x2 + y2 = 8

Y 36. Let the equation of the circle C be
x2 + y2 + 2gx + 2fy + c = 0
A …(i)
Given circle is x2 + y2 = 1 …(ii)
fi x2 + y2 – 1 = 0

Since two circles are orthogonal, so

B 2(g ◊ 0 + f ◊ 0) = c – 1
X¢ O
P(4, 0) X fi c=1

Also, the equation of radical axis is

B 2gx + 2fy + c + 1

Y¢ fi 2gx + 2fy + 1 + 1 = 0

Let the point A be (h, k). fi gx + fy + 1 = 0
The equation of the tangent at A is hx + ky = 8 which is
passing through P(4, 0). It is given that the radical axis parallel to y-axis, so,
So, 4h = 8 fi h = 2
f = 0, g ΠR Р{0}

Now PA = length of the tangent = 2 2 Thus, x = - 1
g

fi PA2 = 8 Hence, the equation of the circle is
fi (h – 4)2 + k2 = 8 fi x2 + y2 + 2gx + 1 = 0
fi (2 – 4)2 + k2 = 8 fi x2 + y2 + x + 1 = 0
fi k2 = 8 – 4 or x2 + y2 – x + 1 = 0

fi k=2 37. Ans.
38. Given circle is (x + r)2 + (y – h)2 = r2.
Hence, the point A is (2, 2).

Now, for the point B, The equation of any tangent passing through origin is

x-2= y-2=4 y = m x. i.e. m x – y = 0
cos q sin q
As we know that the length of the perpendicular from

fi x = 2 + 4 cos q, y = 2 + 4 sin q the centre of the circle to the tangent is equal to the

Thus, the point B is (2 + 4 cos q, 2 + 4 sin q). radius of a circle.
Since the point B lies on the circle x2 + y2 = 8, so
(2 + 4 cos q)2 + (2 + 4 sin q)2 = 8 Thus, - rm - h = r
m2 + 1

fi 16 cos q + 16 sin q + 16 = 0 fi (- rm - h)2 = (r m2 + 1)2

fi cos q + sin q + 1 = 0 fi (rm + h)2 = r2(m2 + 1)
fi r2m2 + 2rmh + h2 = r2m2 + r2
fi cos q + sin q = –1 fi 2rmh + h2 = r2

fi 1 cos q + 1 sin q = — 1
22 2

3.62 Coordinate Geometry Booster

fi m = r2 - h2 and c = 27 – 2f = 27 + 11 = 38
2rh
Hence, the equation of the circle is

x2 + y2 + 7x – 11y + 38 = 0.

Hence, the equation of the circle is 40. Here, the centre = ÁÊË 2, 32¯˜ˆ and the radius = 5
2
y = Ê r 2 - h2 ˆ x
ËÁ 2rh ¯˜
Therefore, the equation of the circle is

fi (r2 – h2)x – 2rhy = 0 ÁËÊ 3 ˆ¯˜ 2 25
2 4
fi (h2 – r2)x + 2rhy = 0 (x — 2)2 + y - =

39. Let the equation of the circle be Y
x2 + y2 + 2gx + 2fy + c = 0
…(i)
Also (i) is orthogonal to …(ii)
x2 + y2 + 4x – 6y + 9 = 0

Thus, 2(g1g2 + f1f2) = c1 + c2 M
fi 2(g ◊ 2 + f ◊ ( –3)) = c + 9 P

fi 4g – 6f = c + 9 C
O
X

C(–g, –f)

P(–2, 7) x+y=5 The equation of the tangent parallel to the diagonal is
A 4y – 3x + k = 0

Now, CM = 5
2

Also, CP ^ PA 4 Ê 3ˆ - 3.2 + k =5
Thus, 7 + f ¥ -1 = -1 ËÁ 2¯˜
fi
g-2 52

fi 7+ f =1 fi k = ± 25
g-2 2

fi 7+f=g–2 …(ii) Hence, the equations of the tangents are
fi f=g–9 …(iii) 4y - 3x ± 25 = 0
From Eqs (i) and (ii), we get 2

4g – 6f = c + 9 fi 8y – 6x ± 25 = 0
fi 4(f + 9) – f = c + 9
fi 36 – 2f = c + 9 41. Clearly, the centre of the circle is Ê -1, - 1ˆ
fi c = 27 – 2f ÁË 4˜¯
Again, CP = Radius
C
-g - f - 5 = g2 + f 2 - c
12 + 12 Q y = 2x + 11
P
fi (g + f + 5)2 = 2(g2 + f 2 – c)
fi (2f + 14)2 = 2((9 + f )2 + f 2 – (27 – 2f)) Given line is y = 2x + 11
fi 2(f + 7)2 = (2f 2 + 18f + 81 – (27 – 2f ))
fi 2(f + 7)2 = (2f 2 + 20f + 54) fi 2x – y + 11 = 0 …(i)
fi (f + 7)2 = (f 2 + 10f + 27)
fi f2 + 14f + 49 = (f 2 + 10f + 27) The equation of CP is –x – 2y + k = 0
fi 4f = 27 – 49
fi f = - 22 = - 11 fi x + 2y – k = 0

42 which is passing through the centre C Ê -1, - 1ˆ . so,
Thus, g = f + 9 = 9 - 11 = 7 -1 - 1 = k ËÁ 4˜¯

22 2

Circle 3.63

fi k=-3 43. If R be the radius of the circumcircle

2 so, R = ph + qk - r …(i)
Hence, the line CP is x + 2y + 3 = 0 p2 + q2
2
fi 2x + 4y + 3 = 0 …(ii) Since TP and TQ are tangents to the circle x2 + y2 =
a2, so the circumcircle through T will pass through its
On solving Eqs (i) and (ii), we get
centre.
x= -9, y=2
2 9 Thus, R = h2 + k 2 ...(ii)
2
Hence, the required point is Ê - , 2¯˜ˆ . From Eqs (i) and (ii), we get
ËÁ

42. Given circle is x2 + y2 + 4x = 0 ph + qk — r = h2 + k 2
p2 + q2
(x + 2)2 + y2 = 4
fi ( ph + qk — r)2 = ( p2 + q2 )(h2 + k 2 )
So, the centre is (–2, 0) and the radius is = 2.

The equation of the line joining the centres of the cir- Hence, the locus of (h, k) is
cles is ( px + qy — r)2 = ( p2 + q2 )(x2 + y2 )

y + 2 = m(x – 0)

y + 2 = 3x, m = tan (60°) = 3 44. Let S1: x2 + y2 – 2x – 6y + 6 = 0
and S2: x2 + y2 + 2x – 6y + 6 = 0
Y S3 : x2 + y2 + 4x + 6y + 6 = 0

T3 Let the equation of the circle passing through the point

of intersection S1 and S2 is

R C2 T2 S4: S1 + lS2 = 0
P
fi (x2 + y2 — 2x — 6y + 6) + l(x2 + y2 + 2x — 6y + 6) = 0

X¢ Q X

C1(2, 0) O T1 fi x2 + y2 + Ê 2l - 2ˆ x + Ê -6l - 6ˆ yz + Ê 6l + 6ˆ = 0
ÁË l +1 ˜¯ ÁË l +1 ˜¯ ËÁ l +1 ˜¯

Y¢ …(i)

Here, let the centre C2 be (h, k). Also, S4 is orthogonal to S3.
Thus, 2(g1g2 + f1f3) = c1 + c2
h + 2 k -0 =4
Thus, 1 = 3
2.2 ÁÊË 2l - 2ˆ 2.3ÊÁË - 6l - 6ˆ Ê 6l + 6ˆ
22 fi l + 1 ˜¯ + l + 1 ˜¯ = 6 + ÁË l + 1 ˜¯

fi h = 0, k = 2 3 On solving, we get
l2 + 6l + 8 = 0
Therefore, C2 = (h, k) = (0, 2 3)
Hence, the locus of the centre of the outer circle is fi (l + 2)(l + 4) = 0
fi l = –2, –4
(x - 0)2 + ( y - 2 3)2 = 4 Put the values of l = –2, –4, we get

x2 + y2 - 4 3y - 8 = 0 x2 + y2 + 6x – 6y + 6 = 0

The equation of the common tangent T1 is or x2 + y2 + Ê 10ˆ x - 6y + 6 = 0
—x - 3y + k = 0 ÁË 3 ¯˜

which is passing through P(-1, 3). Integer Type Questions

So, k = 2 1. Clearly, C1C2 = 5
Hence, the equation of the common tangent T1 is Here, r1 = 10, r2 = 5
x + 3y = 2 . Thus, C1C2 = 5 = r1 – r2
Clearly, the co-ordinates of Q and R are So, the number of common tangents = 1

( 3 - 1, 3 - 1) and (- 3 - 1, 3 + 1) 2. We have, 2(g1g2 + f1f2) = c1 + c2
fi 2(1.0 + k ◊ k) = k + 6
Hence, the equation of the tangents T2 and T3 are fi 2k2 – k – 6 = 0
y - ( 3 - 1) = 3[x - ( 3 - 1)]

and y - ( 3 - 1) = 3[x + ( 3 + 1)]

3.64 Coordinate Geometry Booster

fi (k – 2)(2k + 3) = 0 fi l ◊ 1 = (–1)(–2)
fi k = 2, –3/2 fi l=2
3. Let the equation of the circle be
8.
x2 + y2 + 2gx + 2fy + c = 0
A(0, 6)

and ÁËÊ m, 1 ˜ˆ¯ be a variable point lies on the circle
m

Thus, m2 + 1 + 2gm + 2f +c=0 B(–2 3, 0) C(0, 2)
m2 m B(2 3, 0)

fi m4 + 1 + 2gm3 + 2fm + cm2 = 0 Clearly, radius = 4.
fi m4 + 2gm3 + cm2 + 2fm + 1 = 0 9. Given circle is

It has four roots, say m1, m2, m3 and m4. x2 + y2 – 4x – 6y + l = 0
Thus, m1m2m3m4 = 1 fi (x - 2)2 + ( y - 3)2 = ( 13 - l )2
fi m1m2m3m4 + 4 = 5
4. The equation of the common chord is Clearly, 13 - l = 3

6x + 14y + p + q = 0 fi 13 – l = 9
fi l=4
Here, the centre of the 1st circle (1, –4) lies on the com- 10. Clearly, m1m2 = 1
mon chord. fi m1m2 + 4 = 5

So, 6 – 56 + p + q = 0 Previous Years’ JEE-Advanced Examinations
p + q = 50

Hence, the value of ÁËÊ p+q + 2ˆ¯˜ =7 1. Given lines are 3x + 5y – 1 = 0,
10 (2 + c)x + 5c2y – 1 = 0

5. We have, |r1 – r2| < C1C2 < r1 + r2 On solving, we get
fi |r – 3| < C1C2 < r + 3
fi |r – 3| < 5 < r + 3 fi x = y = 1
- 5 + 5c2 - (2 + c) + 3 5(2
fi –5 < (r – 3) < 5, r + 3 > 5 15c 2 - + c)

fi –2 < r < 8, r > 2 x = y = 1
5(c2 - 1) - (c - -
fi 2<r<8 1) 5(3c2 c - 2)

Clearly, n = 2, m = 8 (c2 - 1) - (c - 1)
(3c2 - c - 5(3c2 - c -
Hence, the value of (m – n) is 6. fi x = 2) , y = 2)

6. The equation of any line passing through P is

y — 2 2 = -(x + 2 2) fi x = (c - 1)(c + 1) , y = - (c - 1)
fi y = –x (c - 1)(3c + 2) 5(c - 1)(3c + 2)

and the equation of the circle is fi x = (c + 1) , y = - 1
(3c + 2) 5(3c + 2)
x2 + y2 = 16
when c tends to 1, then
Y x= 2, y = - 1
P(–2 2, 2 2) 5 25

A Now, radius,
X¢ O
X 2 2

B r= ËÁÊ 2 - 2 ¯ˆ˜ + ÊËÁ 0 + 1 ˜¯ˆ
5 25

Y¢ = 64 + 1 = 1601
25 625 625
Clearly, AB = 8
Hence, the equation of the circle is
7. As we know that if the lines a1x + b1y + c1 = 0 and
a2x + b2y + c2 = 0 cut the x and y axes in four concyclic Ê x - 2ˆ 2 + Ê y + 1 ˆ2 = 1601.
points, then ËÁ 5 ˜¯ ÁË 25˜¯ 625

a1a2 = b1b2

Circle 3.65

2. The equation of a circle passing through the point of fi 9y2 + 60y + 100 + 16y2 –24y – 80 + 64y – 320 = 0
intersection of x2 + y2 = 6 and x2 + y2 – 6x + 8 = 0 is fi 25y2 + 100y – 300 = 0
fi (x2 + y2 – 6) + l(x2 + y2 –6x + 8) = 0). fi y2 + 4y – 12 = 0
which is passing through (1, 1). fi (y + 6)(y – 2) = 0
So, (2 – 6) + l(10 – 6) = 0 fi y = 2, –6
fi l=1 when y = 2, –6, then x = 4, –2
Hence, the equation of the circle is Hence, the points of intersection are (4, 2), (–2, –6)
(x2 + y2 – 6) + 1 ◊ (x2 + y2 – 6x + 8) = 0
fi 2(x2 + y2) – 6x + 2 = 0 7. Equation of any circle passing through the point of in-
fi (x2 + y2) – 3x + 1 = 0
tersection of the given circles is
3. The equations of the tangents at B and D are (2x2 + 2y2 + 4x – 7y – 25) +l(x2 + y2 + 13x – 3y) = 0
y = 7 and 3x – 4y = 20 which is passing through the point (1, 1).
fi –24 + 12l = 0
B(1, 7) fi l=2
Hence, the equation of the circle is
A C
(1, 2) (2x2 + 2y2 + 4x – 7y – 25)
+ 2(x2 + y2 + 13x – 3y) = 0
D(4, –2)
fi 4x2 + 4y2 + 30x – 13y – 25 = 0
8. Ans. (b)

9. Do yourself
10. Equation of the chord bisected at M is

ax + by = a2 + b2

Thus, the co-ordinates of C are (16, 7).

Hence, the area of the quadrilateral ABCD = 2(DABC)

= 2 ËÁÊ 1 ¥ AB ¥ BC˜¯ˆ O
2
= AB ◊ BC
P(h, k) A M(a, b) B
= (5 ¥ BC)

= 5 ¥ 15 which is passing through P(h, k).
So, ah + bk = a2 + b2
= 75 sq. u.
Hence, the locus of M(a, b)
4. hx + ky = x2 + y2

P(–4, 3) x+y=4 11. Let AB be a chord of the circle.
C(h, k) N
Draw a perpendicular from the centre O to the chord

AB at M. Then AM = BM.
Now, –AOM = 45°

M x–y=2 fi cos 45° = OM = OM O
OA 2

Clearly, CM = CN fi OM = 1 A M(h, k) B
fi h+k-4 = h-k-2 22

22 fi OM2 = 2
fi h2 + k2 = 2
fi (h + k – 4) = ±(h – k – 2)
fi h = 3, k = 1 Hence, the locus of M(h, k) is
x2 + y2 = 2
Now, r = (3 + 4)2 + (1 - 3)2 = 53
12. Given equation is x2 = 2ax – b2 = 0
Hence, the equation of the circle is
Let its roots are x1 and x2.
(x – 3)2 + (y – 1)2 = 53 Thus, x1 + x2 = 2a, x1 ◊ x2 = –b2
Similarly, y1 + y2 = –2p, y1 ◊ y2 = –q2
fi x2 + y2 – 6x –2y– 43 = 0
Hence, the equation of the circle is
5. Ans. k π 1
fi (x – x1)x – x2) + (y – y1)(y – y2) = 0
6. Given circle is fi x2 – (x1 + x2)x + x1x2 + y2 – (y1 + y2)y + y1y2 = 0
fi x2 + 2ax – b2 + y2 + 2py – q2 = 0
x2 + y2 – 2x + 4y – 20 = 0
fi x2 + y2 + 2ax + 2py – (b2 + q2) = 0
ÊÁË 3y + 10 ˆ¯˜ 2 2ÁËÊ 3y + 10 ¯ˆ˜
4 4
fi + y2 - + 4y - 20 = 0

fi (3y + 10)2 + 16y2 – 8(3y + 10) + 64y – 320 = 0

3.66 Coordinate Geometry Booster
17. Let the co-ordinates of M be (h, k).
13. Diameter = Distance between two parallel lines

Ê 4 - Ê - 7ˆ ˆ M(h, k)
ËÁ ÁË 2¯˜ ¯˜
=
32 + 42
B
=3
2 A(0, 3)

Hence, the radius is 3 . Clearly, B is the mid-point of AM.
4
Thus, ÁÊË h , k + 3˜¯ˆ
2 2
14. Equation of the chord bisected at M(h, k) is T = S1

Since B lies on x2 + 4x + (y – 3)2 = 0. So

Ê hˆ 2 + 4 Ê hˆ + Ê k + 3 - 2˜¯ˆ 2 = 0
ÁË 2 ¯˜ ËÁ 2 ¯˜ ËÁ 2
C(1, 0)

O M(h, k) B fi h2 + 8h + (k – 3)2 = 0

Hence, the locus of M (h, k) is

fi xx1 + yy1 - (x + x1) = x12 + y12 - 2x1 x2 + 8x + (y – 3)2 = 0
fi hx + ky – (x + h) = h2 + k2 – 2h
18. Since the lines 5x + 12y = 10 and 5x – 12y = 40 touch
which is passing through origin.
So, 0 + 0 – 0 – h = h2 + k2 – 2h the circle C1, its centre lies on one of the angle bisec-
fi h2 + k2 – h = 0 tors of the given lines.

5x + 12y — 10 = 5x - 12y — 40
25 + 144 25 + 144
Hence, the locus of M(h, k) is
x2 + y2 – x = 0. D(4, 5)

15.

B

5x – 12y = 40 C
3
C
A 5x + 12y = 10
B
A 4M

Here, C = (2, 1) and r = 4 fi |5x + 12y – 10| = |5x – 12y – 40|
\ Area of quadrilateral ACBD = 2(DACD) fi (5x + 12y – 10) = ±(5x – 12y – 40)
fi x = 5 and y = –5/4
= 2¥1¥4¥2 Since the centre lies on the first quadrant, let its co-
2 ordinates be (5, k).
We have CM = 3
= 8 sq. u.
16. The equation of the line of intersection is fi 5.5 + 12k - 10 = 3
25 + 144
ÊÁË x2 + y2 - 2 x + 4y - 3¯ˆ˜
3 fi 15 + 12k = 3
13
fi –(x2 + y2 + 6x + 2y – 15) = 0
fi (15 + 12k) = ±39
fi - Ê 2 + 6ˆ¯˜ x + (4 - 2) y + 12 = 0 fi k = 2, - 9
ÁË 3
2
fi - 20 x + 2 y + 12 = 0 Thus, k = 2 and r = 32 + 42 = 5
3
Hence, the equation of the circle is
fi - 10 x + y + 6 = 0 (x – 5)2 + (y – 2)2 = 25
3
fi x2 + y2 – 10x – 4y + 4 = 0
fi –10x + 3y + 18 = 0
fi 10x – 3y – 18 = 0.

Circle 3.67
19.
21. The equation of the chord of contact is
Y 4x + 3y = 9

\ Length of the tangent PQ = PR

= 16 + 9 - 9 = 4

Q M(h, k) Now, QM = 16 - 256 = 12
25 5
S Q

X¢ R O P X
L1 = 0
L2 = 0

Y¢ P(4, 3) M

20. Given circle is

x2 + y2 – 4x – 4y + 4 = 0

fi (x – 2)2 + (y – 2)2 = 4 R

B(0, b) Hence, the area of the triangle PQR
= 2(DPQM)
M = 2 ¥ 1 ¥ 12 ¥ 16
2 25 5
= 192
C(2, 2) 25

O A(a, 0) 22.

Let AB: x + y = 1 C
ab

Now, CM = 2

2 + 2 -1 q
9

fi a b =2 A MB
1 1
a2 + b2 Here, AB = 2. So, AM = 1

2 + 2 -1 In DACM, sin ÊÁË p ˜ˆ¯ = AM = 1
fi a b = ±2 9 AC r

1 + 1 fi r = 1 = cosec Ê p ˆ
a2 b2 ÁË 9 ˜¯
sin Ê p ˆ
2 + 2 -1 ÁË 9 ¯˜

fi a b = -2 23.

1 + 1
a2 b2

[since the origin and (2, 2) lie on the same side of the 3
O¢
line AB] O
5 4
Here, circumcentre = M = ËÊÁ a , b ¯˜ˆ C1
2 2

Therefore the locus of the circumcentre is C2 A

fi 1 + 1 -1= -2 1 + 1 Clearly, OO¢ = 25 — 16 = 3
xy 4x2 4y2 It is given that, tan q = 3

fi 1 + 1 -1= - 1 + 1 4
xy x2 y2 fi sin q = cos q = 1

fi x + y - xy + x2 + y2 = 0 3 45
Thus, the value of k is 1. Let the co-ordinates of the centre O¢ = (x, y) of the cir-
cle C2.

3.68 Coordinate Geometry Booster

x - 0 y - 0 Equation of PQ is
4 3
Therefore, = = ± 3 y — k = — h (x - h)
k
55

fi x = ± 12 , y = ± 9 fi hx + ky = h2 + k2
55
hx + ky
ËÊÁ 12 95ˆ˜¯ fi h2 + k2 = 1 …(i)
5
Thus, O¢ = ± , ± .

The equation of the pair of lines joining the point of

24. The equation of a circle be intersection of (i) with S = 0 is

x2 + y2 + 2gx + 2fy + c = 0 …(i) Ê hx + ky ˆ Ê hx + ky ˆ 2
ÁË h2 + k 2 ¯˜ ÁË h2 + k 2 ¯˜
which passes through (a, b). So x2 + y2 + (2gx + 2fy) + c = 0
a2 + b2 + 2ga + 2fb + c = 0
…(ii) …(ii)
Also (i) is orthogonal to x2 + y2 = k2. So
2(g ◊ 0 + f ◊ 0) = c – k2 As the line (ii) are at right angles, so

fi c = k2 …(iii) 1+1+ Ê 2gh + 2fk ˆ + c (h2 + k2) = 0
ÁË h2 + k 2 ˜¯ (h2 + k2)2
From Eqs (ii) and (iii), we get
fi (h2 + k 2 ) + gh + fk + c = 0
a2 + b2 + 2ga + 2fb + k2 = 0 2
fi a2 + b2 – 2(–g)a – 2(–f)b + k2 = 0

Hence, the locus of the centre (–g, –f) is Hence, the locus of M(h, k) is
a2 + b2 – 2xa – 2yb + k2 = 0 fi (x2 + y2 ) + gx + fy + c = 0

fi 2ax – 2by – (a2 + b2 + k2) = 0 2
27. Given two circles intersect. We have
25. Given circle is (x + r)2 + (y – h)2 = r2

Equation of any tangent passing through origin is |r1 – r2| < C1C2 < r1 + r2
fi |r – 3| < 5 < r + 3
y = mx fi mx – y = 0 fi |r – 3| < 5, 5 < r + 3
fi –5 < (r – 3) < 5, r > 2
As we know that the length of the perpendicular from fi –2 < r < 8, r > 2
fi 2<r<8
the centre of the circle to the tangent is equal to the

radius of a circle.

Thus, - rm - h = r 28. Let r be the radius of the circle.
m2 + 1

fi (- rm - h)2 = (r m2 + 1)2 So, pr2 = 154

fi (rm + h)2 = r2(m2 + 1) fi 22 ¥ r2 = 154
fi r2m2 + 2rmh + h2 = r2m2 + r2 7
fi 2rmh + h2 = r2
fi r2 = 7 ¥ 7

fi m = r2 - h2 fi r=7
2rh
Now, the centre of the circle is the point of intersection

of 2x – 3y = 5 and 3x – 4y = 7. So

Hence, the equation of the circle is C = (1, –1)

y = Ê r 2 - h2 ˆ x Hence, the equation of the circle is
ÁË 2rh ˜¯
(x – 1)2 + (y + 1)2 = 72
fi (r2 – h2)x – 2rhy = 0 fi x2 + y2 – 2x + 2y – 47 = 0
fi (h2 – x2)x + 2rhy = 0
29 Let the equation of the circle be

26. Let PQ be the chord of the given circle which subtends x2 + y2 +2gx + 2fy + c = 0

a right angle at the origin O and M(h, k) be the foot of Let ÊËÁ m, 1 ¯ˆ˜ be an arbitrary point on the circle.
m
the perpendicular from O on this chord PQ.

P fi m2 + 1 + 2gm + 2f +c=0
m2 m

fi m4 + 1 + 2gm3 + 2fm + cm2 = 0

O M(h, k) fi m2 + 2gm3 + cm2 + 2fm + 1 = 0
Q
Let m1, m2, m3, m4 be the roots.

Thus, m1 ◊ m2 ◊ m3 ◊ m4 = 1 =1.
1

Circle Also, r = CM 2 + (3 2)2 3.69
…(iii)
30. Given circle is x2 + y2 – 6x + 2y = 0 fi r = CM 2 + 18
Centre is (3, –1).
Clearly, the centre (3, –1) satisfies the equation of the fi r2 = CM2 + 18
diameter x + 3y = 0.
fi r2 = (h + k)2 + 18
31. 2

Y fi (h — k)2 = (h + k)2 + 18
22
B
fi (h — k)2 - (h + k)2 = 18
P 22

X¢ O M A X

fi –2hk = 18

fi hk = –9

Y¢ From Eq. (ii), we get

The equation of tangent to the circle x2 + y2 = 4 at h+k=8
(1, 3) is fi h— 9 =8

x + 3y = 4 h
The equation of normal to the circle x2 + y2 = 4 at fi h2 – 8h – 9 = 0
(1, 3) is fi (h – 9)(h + 1) = 0
fi h = –1, 9
y = 3x
and k = 9, –1
Thus, area of the triangle OPA
= 1 ¥ OA ¥ PM Hence, the equation of the circle is
2 (x + 1)2 + (y – 9)2 = 50
=1¥4¥ 3
2 or
= 2 3 sq. u. (x – 9)2 + (y + 1)2 = 50

32. Let C(h, k) be the centre and r be the radius of the given 33. Ans. 3 3 ¥ r2 sq. units
circle. 4

Y 34.

4x + 3y = 10

55
C1 P (1, 2) C2

X¢ O Q X The equation of the line where both the centres lie is
P M 3x – 4y + k = 0 which is passing through the point.
R Thus, k = 5
x–y=0 C x+y=0 Hence, the required line is 3x – 4y + 5 = 0
Clearly, tan q = 3
Y¢
4
Thus, h - k = r …(i)
2 …(ii) fi sin q = cos q = 1
3 45
Given OP = 4 2, QR = 6 2
Therefore, the co-ordinates of C1 and C2 can be ob-
So, QM = MR = 3 2 tained from

Clearly, CM = OP = 4 2 fi x - 1 = y-2 = ±5
4 3

h+k =4 2 55
2
fi x = 1 ± 4, y = 2 ± 3
fi x = 5, –3; y = 5, –1

3.70 Coordinate Geometry Booster

Therefore, the equations of the required circles are 37. Given circle is
(x – 5)2 + (y – 5)2 = 25 2x(x – a) + y(2y – b) = 0

or Y
(x + 3)2 + (y + 1)2 = 25
P(a, b/2)
35. Here, A, B, C and D are concyclics.

D(0, 1) O M(c, 0) X
lx – y + 1 = 0

– 2y + 3= 0 B(0, 3/2) fi 2x2 – 2ax + 2y2 – by = 0
O fi 2x2 + 2y2 – 2ax – by = 0
x

A(–3, 0) C – 1 , 0
l
The equation of the chord bisected at M is T = S1

Thus, OA.OC = OB.OD fi 2xx1 + 2yy1 - a(x + x1 ) - b Ê y + y1 ˆ
ËÁ 2 ¯˜
fi - 3. - 1 = 3 ◊1
l2 = 2x12 + 2y12 - 2ax1 - by1

fi l=2 fi 2x◊c + 0 - a(x + c) - b ÁÊË y + 0¯ˆ˜
2
36. Consider three circles with centres at A, B and C with = 2c2 + 0 – 2ac – 0
radii r1, r2, r3 respectively, which touch each other ex-
ternally at P, Q, R. fi 4cx + 2ax – 2ac – by = 4c2 – 4ac

which is passing through P(a, b/2). So

C 4ca - 2a2 — 2ac - b2 = 4c2 - 4ac
r3 r3 2

P R fi 8ca – 4a2 – 4ac – b2 = 18c2 – 8ac
r1 O r2 fi 8c2 – 12ac + (4a2 + b2) = 0
A
r1 Q r2 B
Now, D > 0
Let the common tangents at P, Q, R meet each other at fi 144a2 – 32(4a2 + b2) > 0
fi 9a2 – 2(4a2 + b2) > 0
O. fi 9a2 + 8a2 – 2b2 > 0
fi a2 > 2b2
Then OP = OQ = QR = 4
38.

Y

Also, OP ^ AB, OQ ^ AC, OR ^ BC

Here, O is the incentre of the triangle ABC.

For DABC, C(h, k)

s = (r1 + r2 ) + (r3 + r2 ) + (r1 + r3 ) = r1 + r2 + r3 X¢ X
2
O A(1, 0)

and D = (r1 + r2 + r3)r1r2r3 Y¢
Now, from the relation r = D , we get
Clearly, OC = OA
s fi OC2 = CA2
(r1 + r2 + r3)r1r2r3 = 4 fi (h2 + k2) = (h – 1)2 + k2
fi (h2 + k2) = h2 – 2h + 1 + k2
r1 + r2 + r3 fi –2h + 1 = 0

fi r1r2r3 = 4 fi h=1
r1 + r2 + r3 2

fi r1r2r3 = 16 = 16
r1 + r2 + r3 1

fi (r1r2r3) : (r1 + r2 + r3) = 16 : 1

Circle 3.71

Also, C1C2 = |r1 – r2| fi –5x – 6y + 56 = 0, 2x + 3y – 27 = 0
fi OC = 3 - h2 + k 2
fi x = 2, y = 23
3
Ê 23ˆ
fi h2 + k2 = 3 - h2 + k2 Hence, the co-ordinates of the point is ÁË 2, 3 ¯˜ .

fi 2 h2 + k2 = 3 41. Given circles are

fi h2 + k2 = 9 x2 + y2 – 4x – 2y + 4 = 0
4
fi (x – 2)2 + (y – 1)2 = 1

fi k2 = 9 - h2 = 9 - 1 = 2 and (x – 6)2 = (y – 4)2 = 42
4 44
fi (x – 6)2 + (y – 4)2 = 42

fi k= 2 Here, C1 = (2, 1), r1 = 3; C2 = (6, 4), r2 = 4

Hence, the centre is Ê 1 , 2¯˜ˆ . Now, C1C2 = (6 - 2)2 + (4 - 1)2 = 5
ÁË 2
= r1 + r2
39. Given circle is
x2 + y2 – 6x – 6y + 14 = 0 N
M
fi (x – 3)2 + (y – 3)2 = 4
P 14
C1(2, 1) C2(6, 1)
Y

C1(h, k) C2(3, 3) Therefore, D = Ê 4.2 + 1.6 , 4.1 + 1.4ˆ = ËÊÁ 14 , 85¯ˆ˜
X ËÁ 4 + 1 4 + 1 ˜¯ 5

O and P = Ê 4.2 - 1.6 , 4.1 — 1.4ˆ = Ê 2 , 0˜ˆ¯
ËÁ 4 - 1 4 - 1 ˜¯ ÁË 3

The equations of the tangent through P is

fi y - 0 = m ËÊÁ x - 2 ¯˜ˆ
3
We have,
C1C2 = r1 + r2 fi 3y = m(3x – 2)

fi (h - 3)2 + (k - 3)2 = h + 2 fi 3mx – 3y – 2m = 0

fi (h – 3)2 + (k – 3)2 = (h + 2)2 Now, C1M = 1
fi h2 – 6h + 9 + k2 – 6k + 9 = h2 + 4h + 4
fi k2 – 10h – 6k + 14 = 0 fi 3m.2 - 3.1 - 2m = 1
Hence, the locus of (h, k) is 9m2 + 9

y2 – 10x – 6y + 14 = 0 fi 4m - 3 = 1
3 m2 + 1
40. The equation of any circle passing through A(3, 7) and
B(6, 5) is fi (4m – 3)2 = 9(m2 + 1)
fi 16m2 – 24m + 9 = 9m2 + 9
x y1 fi 7m2 – 24m = 0
(x - 3)(x - 6) + ( y - 7)( y - 5) + l 3 7 1 = 0 fi m(7m – 24) = 0
fi m = 0, (7m – 24) = 0
6 51 fi m = 0, m = 24

fi S1:(x – 3)(x – 6) + (y – 7)(y – 5) + l(2x + 3y – 27) = 0 7

fi S1: x2 + y2 – 9x – 12y + 53 + l(2x + 3y – 27) = 0 Hence, the equations of tangents are
…(i)
y = 0 and y = 24 (x - 3)
The equation of the common chord of (i) 7

and S2: x2 + y2 – 4x – 6y – 3 = 0 …(ii) fi y = 0 and 24x – 7y – 72 = 0

is S1 – S2 = 0. 42. Given circle is
fi –5x = 6y + 56 + k(2x + 3y – 27) = 0 4x2 + 4y2 – 12x + 4y + 1 = 0
x2 + y2 – 3x + y + (1/4) = 0
x2 + y2 – 3x + y + (1/4) = 0

3.72 Coordinate Geometry Booster

Ê 3 1ˆ 9+1-1 =3 Let x be the side of a square.
ËÁ 2 2¯˜ 444 2
Thus, C = , - , r = Ê aˆ 2 a2
ÁË 3 ˜¯ 3
Thus, x2 + x2 = =

In DACM,

cos (60°) = CM fi 2x2 = a2
AC 3

fi CM = 1 C(3/2, 1/2) fi x2 = a2
3/2 2 6
a2
fi CM = 3 3/2 60°
4 A M(h, k) B Area of a square =
6
fi CM 2 = 9
16 45. Given circle is
x2 + y2 + 4x – 6y + 9 sin2 a +13 cos2 a = 0
ÊÁË h 3 ˜¯ˆ 2 ÁÊË k 1 ˜¯ˆ 2 9
2 2 16 (x + 2)2 + (y – 3)2 = 4 – 4 cos2 a
fi - + - =
(x + 2)2 + (y – 3)2 = (2 sin a)2
Hence, the locus of M(h, k) is
(x + 2)2 + (y – 3)2 = (2 sin a)2
ÊÁË 3 ˆ¯˜ 2 Ê 1ˆ 2 9
2 ËÁ 2˜¯ 16 Let the P be (h, k).
x - + y - =

A

43. Here, C1 = (5, 0), r1 = 3, C2 = (0, 0), r2 = r P(h, k) a C(–2, 3)
It is given that two circles intersect.
B
So, |r1 – r2| < C1C2 < r1 + r2
fi |3 – r| < 5 < 3 + r We have
fi |3 – r| < 5, 5 < 3 + r
fi –5 < (r – 3) < 5, r > 2 sin a = AC
fi –2 < r < 8, r > 2 PC
fi 2<r<8
fi sin a = 2 sin a
44. Let PM = p
(h + 2)2 + (k - 3)2

P fi (h + 2)2 + (k – 3)2 = 4
fi h2 + k2 + 4h – 6k + 9 = 0

D C Hence, the locus of (h, k) is
O x2 + y2 + 4x – 6y + 9 = 0

46. Let r be the radius of a circle, then AC = 2r

AB
QMR

Now, sin (60°) = PM = p
PQ a

fi p= 3
a2

fi p= 3a
2

Here O is the centroid. Since, AC is the diameter –ABC = 90°
In DABC, BC = 2r sin b, AB = 2r cos b
So the centroid divides the median in the ratio 2 : 1.
BD = AB tan a = 2r cos b tan a
Thus, OM = p AD = AB sec a = 2r cos b sec a
3 DC = BC – BD = 2r sin b – 2r cos b tan a
Since E is the mid-point of DC , so
fi r= p DE = DC = r sin b - r cos b tan a
3
2
fi 2r = 2p = 2 ¥ 3 a = a
332 3

Circle 3.73

Now, in DADC , AE is the median Hence, the required equation of the circle is
x2 + y2 – 2x + (x – y) = 0
2( AE2 + DE2 ) = AD2 + AC2
fi x2 + y2 – x – y = 0
2(d 2 + r2 (sin b - cos b tan a )2 )
49. Given C is the circle with centre at (0, 2)
= 4r2 cos2b sec2a + 4r2 and radius r (say)
Then x2 + ( y - 2)2 = r2
r2 = cos2a + cos2b d 2 cos2a b cos(b - a)
+ 2 cos a cos (y - 2)2 = r2 - x2

Hence, the area of a circle = pr2 (y - 2) = ± r2 - x2
p d 2 cos2a
y = 2 ± r2 - x2
cos2a + cos2b + 2 cos a cos b cos (b - a )

Ê1+ a 2 ,1- a 2ˆ The only rational value of y is 0
ÁË 2 2 ˜¯
47. Let (p, q) = Suppose the possible value of x for which y is 0 is x1.
Certainly, y – x1 will also give the value of y as 0. Thus,
Given circle is 2(x2 + y2) – 2px – 2qy = 0 atmost there are two rational points which satisfy the

A(p, q) equation of the circle.

50. Let the point P be (h, k).

M(h, –h) Thus, x - p = y - q = r …(i)
B x+y=0 cos q sin q

fi x = p + r cos q, y = q + r sin q

The point P lies on the curve.
So, a(p + r cos q)2 + b(q + r sin q)2

The equation of the chord bisected at M is + 2h(p + r cos q)(q + r sin q) = 1

2(hx – hy) – p(x + h) – q(y – h) fi (a2 cos2 q + 2h cos q sin q + b2 sin2 q) r2
= 2(h2 + h2) – 2ph + 2qh
+2[p(a cos q + h sin q) r + q(h cos q + b sin q)]r

which is passing through A(p, q). + ap2 + 2hpq + bq2 – 1 = 0 …(ii)
fi 2(hp – hq) – p(p + h) – q(q – h) = 4h2 – 2ph + 2qh
fi 3hp – 3hq – p2 – q2 – 4h2 = 0 Let PQ = r1 and PR = r2.
fi 4h2 + 3(q – p)h + (p2 + q2) = 0
Also, let r1 and r2 are the roots of Eq. (ii)

Thus, r1r2 = (a 2(ap2 + 2hpq + bq2 - 1) 2q
+ b) + 2hsin 2q + (a - b) cos
Since chords are distinct, so

D>0 Since the product of PQ and PR is the independent of q
fi 9(q – p)2 – 16(p2 + q2) > 0 so, h = 0, a = b and a π 0.
So the given product becomes
fi 9(-a 2)2 - 8(1 + 2a2 ) > 0
fi 18a2 – 16a2 – 8 > 0 x2 + y2 = 1
fi 2a2 – 8 > 0 a
fi a2 – 4 > 0
which represents a circle.
fi (a + 2)(a – 2) > 0
51.
fi a < –2, a > 2
52. Any point on the line 2x + y = 4 can be considered as
Thus, a Œ (– , 2) » (2, ) P(a, 4 – 2a).
The equation of the chord of contact of the tangent to
48. The equation of any circle passing through the point of the circle x2 + y2 = 1 from P is
intersection of x2 + y2 – 2x = 0 and y = x is xx1 + yy1 – 1 = 0
fi ax + (4 – 2a)y – 1 = 0
x2 + y2 – 2x + l(x – y) = 0 fi a(x – 2y) + (4y – 1) = 0

fi x2 + y2 + (l – 2)x – ly = 0 fi (x - 2y) + 1 (4y - 1) = 0
a
Its centre is Ê2 —l, lˆ
ÁË 2 2 ¯˜ fi (x - 2y) + l(4y - 1) = 0, l = 1
a
The centre lies on y = x
Thus, x – 2y = 0, 4y – 1 = 0

So, 2 — l = l fi x= 1, y= 1 .
22 24
the required Ê 1 1 ˆ
fi l=1 Hence, point is ÁË 2 , 4 ¯˜ .

3.74 Coordinate Geometry Booster

53. Since two vertices of an equilateral triangle are h = 2x
B(–1, 0) and C(1, 0).
So, the third vertex must lie on the y-axis. Similarly, k = 2y
Let the third vertex be A(0, b)
Now, AB = BC = CA Putting in (i), we get,
fi AB2 = BC2 = AC2 4x2 + 4y2 = 4r2
fi 1 + b2 = 4 = 1 + b2 x2 + y2 = r2
fi b2 = 4 – 1
Hence, the locus is x2 + y2 = r2
fi b= 3
56. Here A0A1 = 1
Thus, the third vertex is A = (0, 3). fi cos (120°) = 12 + 12 - A0 A22
2.1.1
As we know that in case of an equilateral triangle,
fi - 1 = 12 + 12 - A0 A22
Ê 1ˆ 22 A3
ËÁ 3 ˜¯ O A2
Circumcentre = Centroid = 0, fi A0A22 = 3 A4
A1
Hence, the equation of the circumcircle is fi A0 A2 = 3
Similarly, A0 A4 = 3
Ê 1 ˆ 2 Ê 1 ˆ2 Hence, the value of
ÁË 3 ˜¯ ÁË 3 ¯˜
( x — 0)2 + y - = (1 - 0)2 + 0 - A5

fi x2 + Ê y - 1 ˆ2 = 4 . A0 A1 A0 A2 A0 A4 A0
ÁË 3 ˜¯ 3
= 1◊ 3 ◊ 3
54. Given circles are x2 + y2 = 4 = 3.
and x2 + y2 – 6x – 8y – 24 = 0
57. Y

Here C1 = (0, 0), r1 = 2 ; C2 = (3, 4), r2 = 7 B(p, q)
Now, C1C2 = 5 = r2 – r1
So, two circles touch each other internally.

Thus, the number of common tangents = 1

55. Let P(h, k) be on C2 ...(i) O M(h, 0) X
So, h2 + k2 = 4r2 A

The equation of the chord bisected at M(h, 0) is

xx1 + yy1 - p ÊÁË x + x1 ˆ˜¯ - q ÁÊË y + y1 ˆ¯˜
2 2

= x12 + y12 - px1 - qy1

fi hx + 0 - p ÊÁË x + h ˜ˆ¯ - q ËÁÊ y + 0ˆ¯˜
2 2

= h2 + 0 – ph – 0
fi 2hx – px – ph – qy = 2h2 – 2ph

Chord of contact of P w.r.t C1 is hx + ky = r2 fi 2hx – px – qy = 2h2 – ph
It intersects C1 x2 + y2 = a2 in A and B. fi 2h2 – 2hx – ph + (px + qy) = 0
Eliminating y, we get,
which is passing through (p, q)
Ê r2 - hx ˆ 2 So, 2h2 – 2ph – ph + (p2 + q2) = 0
ÁË k ˜¯ fi 2h2 – 3ph + (p2 + q2) = 0
x2 + = r2
Clearly, D > 0

(h2 + k 2 )x2 - 2r2hx + r2 (r2 - k 2 ) = 0 fi 9p2 – 8(p2 + q2) > 0

4r2 x2 - 2r2hx + r2 (r2 - k 2 ) = 0 fi p2 – 8q2 > 0

fi p2 > 8q2

Thus, x1 + x2 = h, y1 + y2 = k 58. The given circle is x2 + y2 = r2 ...(i)
2 2
Centre is (0, 0) and radius = 1

Let (x, y) be the centroid of DPAB Let T1 and T2 be the tangents drawn from (–2, 0) to the
circle (i)
Thus, 3x = x1 + x2 + h = h + h = 3h
2 2

Circle 3.75

Let m be the slope of the tangent, then the equations of For the circles (iii) and (iv), there will be four com-
mon tangents of which 2 are direct and another two are
tangents are transverse common tangents.
In two triangles, DC1XN, DC2YN
y – 0 = m(x + 2)
C1N = 3 = 9
mx – y + 2m = 0 ...(ii) C2N 1/3

Thus, N divides C1C2 in the ratio 9:1

Clearly, 2m = 1 Clearly, N lies on x-axis
m2 + 1

m=± 1 N = Ê - 4 , 0˜ˆ¯
3 ÁË 5

Thus, the two tangents are Any line through N is y = m Ê x + 4ˆ
T1: x + 3 y = 2 ËÁ 5 ˜¯

T2: x - 3 y = 2 5mx – 5y + 4m = 0

Now any other circle touching (i) and T1, T2 is such that If is is tangent to the circle (iii), then
its centre lies on x-axis
Let (h, 0) be the centre of such circle 20m + 4m = 3
Thus, OC1 = OA + AC1 2m2 + 25m

|h| = 1 + |AC1| m=± 5
But AC1 = Perpendicular distance from (h, 0) to the 39
tangents
Hence, the required tangents are
|h| =1+ h + 2
2 y=± 5 Ê x + 4ˆ
39 ÁË 5 ¯˜
|h| -1= h + 2
2 59. Let z1 = Q = 3 + 4i
Then z2 = R = iz = i(3 + 4i) = 3i – 4
h2 - 2 |h| + 1 = h2 + 4h + 4 = – 4 + 3i
4
= (–4, 3)
h = - 4, 4
3 Thus, –QOR = p .
2
Hence, the centres of the circles are
Clearly, –QPR = p .
Ê - 4 , 0˜ˆ¯ , (4, 0) 4
ËÁ 3
60. The given circles are orthogonal. So,

Radius of the circle with centre (4, 0) is 4 – 1 = 3 and the 2(g1g2 + f1f2) = c1 + c2
fi 2(1.0 + k◊k) = k + 6
Ê 4 0¯˜ˆ 4 -1= 1
radius of the circle with centre ÁË - 3 , is 33 fi 2k2 = k + 6

Thus, two possible circles are fi 2k2 – k – 6 = 0

(x – 4)2 + y2 = 9 ...(iii) fi 2k2 – 4k + 3k – 6 = 0

and Ê x + 4ˆ 2 + y2 = 1 ...(iv) fi 2k(k – 2) + 3(k – 2) = 0
ÁË 3˜¯ 9
fi (k – 2)(2k + 3) = 0

Since (i) and (iii) are two touching circles, so they have fi (k – 2) = 0, (2k + 3) = 0

three common tangents T1, T2 and x = 1 fi k = 2, - 3
Similarly, common tangents of (i) and (iv) are T1, T2 2
and x = –1

3.76 Coordinate Geometry Booster
61. 64.

Y C r
B(0, r) P(h, k) R(0, 0)

G(h, k) r1 C1
X¢ O A(r, 0) X
r2 Q(a, b)

P

Y¢ C2

Let A = (r, 0), B = (0, r) Let C1 be the circle with centre R(0, 0) and radius r.
and P = (r cos q, r sin q). Thus, its equation is x2 + y2 = r2.
We have h = 1 (r + r cos q ), k = 1 (r + r sin q ) Let C2 be (x – a)2 + (y – b)2 = r12
and C be (x – h)2 + (y – k)2 = r22.
33 It is given that PR = r – r1 and QR = r + r2

ÊÁË h r ¯ˆ˜ 2 ÁËÊ k 3r ˜ˆ¯ 2 r2 h2 + k 2 = r - r1 and (h — a)2 + (k - b)2 = r1 + r2
3 9
\ — + — = (cos2q + sin2q ) Adding, we get

Hence, the locus of G(h, k) is h2 + k 2 + (h — a)2 + (k - b)2 = r + r2

ÁËÊ 3r ˜ˆ¯ 2 ÁËÊ 3r ¯˜ˆ 2 r2 Thus, the locus of P(h, k) is
9
x — + y — = x2 + y2 + (x — a)2 + ( y - b)2 = r + r2

which represents a circle. which represents an ellipse with foci at R(0, 0) and
Q(a, b) and the length of the major axis is r + r2.
62. Ans. OA = 9 + 3 10 65. Clearly, the point Q is (0, 3).
63. Let –SPR = q

S

Q 5x –2x + 6 = 0 Q(0, 3)
X
X¢ X

P

q Or R Y¢
Pr
Now, the length of the tangent PQ from Q to the circle
Then –QRP = ÁÊË p - qˆ¯˜ , –PQR = q x2 + y2 + 6x + 6y = 2 is
2
PQ = 0 + 9 + 0 + 18 - 2 = 5
In DPQR, tan ËÊÁ p - q¯ˆ˜ = PQ 66. Given common tangent is
2 PR
y = mx - b 1 + m2
fi cot q = PQ …(i)
2r …(ii) fi mx - y - b 1 + m2 = 0

Also, in DPRS, tan q = RS = RS Now, the length of the perpendicular from the 2nd cir-
PR 2r cle is equal to the radius of the circle.

From Eqs (i) and (ii), we get am - b 1 + m2
fi =b
PQ ◊ RS = 1
2r 2r m2 + 1

fi 4r2 = PQ ◊ RS fi am - b 1 + m2 = b m2 + 1

fi 2r = PQ ◊ RS fi am = 2b 1 + m2

Circle 3.77

fi a2m2 = 4b2(1 + m2) 68. From Figure (i),

fi a2m2 = 4b2 + 4b2m2 I = n ◊ 1 ◊ (OA1) ◊ (OA1) sin ËÊÁ 2p ˆ˜¯
2 n
fi (a2 – 4b2) m2 = 4b2 n

fi m2 = 4b2 = p sin ËÊÁ 2p ˆ¯˜
(a2 - 4b2 ) 2 n

4b2 From figure (ii)
(a2 - 4b2 )
fi m= B1B2 = 2(B1L)

fi m= 2b = 2(OL) tan Ê pˆ
ÁË n ˜¯
(a2 - 4b2 )
Ê pˆ
67. Given circle is x2 + y2 = r2 = 2 ◊1 ◊ tan ËÁ n ˜¯

A ÁÊË p ˜ˆ¯
n
= 2 tan

P(6, 8) q q Thus, On = n Ê 1 (B1B2 )(OL)ˆ˜¯ = n tan Êpˆ
MO ÁË 2 ËÁ n ¯˜

B Now, In = (n/2) sin (2q ) ,
On n tan q

We have OP = 62 + 82 = 10 where q = p = 2 tan q ◊ 1 = cos2 q
n (1 + tan2q ) tan
BM = r cos q, OM = r sin q 2 q
where 0 < q < p
= 1 (2 cos2q )
2 2
Also, sin q = r
= 1 (1 + cos 2q )
10 2

If A denotes the area of the triangle PAB, then 1 Ê Ê 2In ˆ 2 ˆ
A = 2ar (DPBM) Á1 ËÁ n ¯˜ ˜
= + 1 - ¯
= 2 ¥ 1 ¥ PM ¥ BM 2Ë
2
On Ê Ê 2In ˆ 2 ˆ
= PM ¥ BM 2 Á1 ËÁ n ˜¯ ˜
= (OP – OM) ◊ BM In = Ë + 1 - ¯
= (10 – r sin q) ◊ r cos q
= (10 – 10 sin q ◊ sin q)(10 sin q ◊ cos q) 69. Given x2 – 8x + 12 = 0 and y2 + 14y + 45 = 0
= 100 sin q cos3 q
fi (x – 2)(x – 6) = 0 and (y – 5)(y – 9) = 0

fi x = 2, x = 6 and y = 5, y = 9

dA = 100 cos q cos3q - 300 sin2q cos2q y=9 C(6, 9)
dq
fi

= 300 cos4q Ê 1 - tan q ˆ Ê 1 + tan q ˆ X=2 x=6
ÁË 3 ¯˜ ÁË 3 ¯˜

For maxima and minima, dA = 0 gives
dq
1 - tan q = 0
3 A(2, 5) y = 5

fi q=p Hence, the centre is Ê2+ 6, 5 + 9ˆ = (4, 7)
6 ËÁ 2 2 ¯˜

dA Ï > 0 : 0 <q < p 70. The equation of a circle C is …(i)
dq ÔÔ <q < 6 …(ii)
Thus, = Ì p (x – 2)2 + (y – 1)2 = r2
ÔÓÔ< p 2
0 : 6 Given circle is
x2 + y2 – 2x – 6y + 6 = 0
Therefore A is maximum, when q=p and
6 The equation of the common chord is
ÊÁË p ¯ˆ˜ –2x + 4y + 5 – r2 – 6 = 0
r = 10.sin 6 = 5
which is a diameter of the circle (ii).

3.78 Coordinate Geometry Booster

Thus, –2.1 + 4.3 + 5 – r2 – 6 = 0 By the rotation theorem,
fi 9 – r2 = 0
fi r=3 Ê z2 - z0 ˆ = Ê z2 - z0 ˆ eip /2 = i
ËÁ z1 - z0 ˜¯ ËÁ z1 - z0 ˜¯
71. Given z - a = k
z-b fi (z2 – z0) = i(z1 – z0)
fi z2 = z0 + i(z1 – z0)
|z - a|2 k2
fi |z - b|2 = fi z2 = 1 + i(2 + i 3 - 1) = 1 + i - 3

fi (z - a )(z - a ) = k 2 fi z2 = (1 - 3) + i
(z - b)(z - b )
Also, z4 = 2 - z2 = 2 — (1 - 3) + i
fi (z - a )(z - a ) = k 2 (z - b)(z - b )
fi z4 = (1 + 3) - i
fi |z|2 - a z - a z + |a|2 = k 2 (|z|2 - b z - b z + |b|2 )
73. The equation of the family of circle is
fi (1 - k 2 ) |z|2 - (a - k 2b)z - (a - b k 2 )z (x – 1)2 + (y + 1)2 + l(2x + 3y + 1) = 0
+ (|a|2 - k |b|2 ) = 0 x2 + y2 – 2x + 2y + 2 + l(2x + 3y + 1) = 0
x2 + y2 + 2 (l – 1) x + (3l + 2) y + (l + 2) = 0
|z|2 (a - k 2b) (a - b k 2 ) …(i)
(1 - k 2 ) (1 - k 2 )
fi - z - z The equation (i) is orthogonal to the circle

+ (|a|2 - k |b|2 ) = 0 x(x + 2) + (y + 1)(y – 3) = 0
(1 - k2) x2 + y2 + 2x – 2y – 3 = 0

Therefore,

Thus, the centre of a circle is = (a - k2b ) . 2 ÈÍÎ(l - 1) ◊1+ (3l + 2) ◊ (-1)˚˙˘ = l -1
(1 - k2) 2
and its radius
fi 2l – 2 – 3l – 2 = l – 1
2
(a - k 2b ) Ê |a|2 - k 2bb ˆ fi l=-3
= (1 - k2) - ÁË (1 - k2) ˜¯ 2

Ê (a - k 2b)ˆ Ê (a - k 2b )ˆ Ê |a|2 - k 2bb Hence, the equation of the circle is
ËÁ (1 - k 2 ) ¯˜ ÁË (1 - k 2 ) ˜¯ ËÁ (1 - k 2 )
= - ˆ (x - 1)2 + ( y + 1)2 — 3 (2x + 3y + 1) = 0
¯˜ 2

k(a - b) fi 2(x2 + y2) – 4x + 4y + 4 – 6x – 9y – 3 = 0
1- k2
= fi 2(x2 + y2) – 10x – 5y + 1 = 0

72. 74. Let A, B, C be the centres of the 3- given circles.

Q(z2) P(2 + i 3)
(z1)
A3
(z0) 3 5 B
O(1, 0) 5
P

4
4

C

R(z3) S(z4)

Since the centre of a square coincides with the centre of Clearly P is the incentre of the DABC.
Thus, r = D = s(s - a)(s - b)(s - c)
a circle, so
ss
z1 + z3 = 1 fi r = (s - a)(s - b)(s - c)
2
s
fi z1 + z3 = 2 fi r = 5.4.3 = 5, since s = 12
fi z3 = 2 — z1 = 2 - (2 + i 3) = - i 3
12
Here, –z1z0 z2 = p
2

Circle 3.79
75. Given C1 = (0, 1)

Y

C r1 r2 C2 X
X¢ O

Y¢

Let C2 = (x, y) PA2 + PB2 + PC2 + PD2
Then r1 + r2 = C1C2 Then QA2 + QB2 + QC2 + QD2

fi 1 + | y| = x2 + ( y - 1)2 = 1+1+ 5 + 5
fi 1 + y2 + 2|y| = x2 + y2 – 2y + 1 2[( 2 - 1)2 + 1] + 2[( 2 + 1)2 + 1]
fi x2 = 2|y| + 2y
fi x2 = 4y if y ≥ 0 = 12 = 3 = 0.75
16 4

Thus, the locus of its centre is 78. Let C ¢ be the said circle touching C1 and L, so that C1
and C ¢ are on the same side of L.
{(x, y) : x2 = 4y} » {(0, y) : y £ 0}
Let us draw a line T parallel to L at a distance equal to
76. The equation of the tangent to the curve y = x2 + 6 at
the radius of the circle C1, on opposite side of L
P(1, 7) is Then the centre of C ¢ is equivalent from the centre of

2x – y + 5 = 0 …(i) C1 and from line T

Y

P(1, 7)

Q X

X¢
C(–8, –6)

Y¢

Here CQ is perpendicular to PQ. Locus of centre of C ¢ is a parabola
79. Since S is equidistant from A and line BD, it traces a
The equation of CQ is –x – 2y + k = 0
parabola.
which is passing through the centre (–8, –6). So,

8 + 12 + k = 0

k = –20

The equation of CQ is –x – 2y – 20 = 0

x + 2y + 20 = 0 …(ii)

On solving Eqs. (i) and (ii), we get

x = –6 and y = –7.

Therefore, the co-ordinates of Q are (–6, –7)

77. Without loss of genrality, we can assume the square Clearly, AC is the axis, A(1, 1) is the focus and
ABCD with its vertices A(1, 1), B(–1, 1), C(–1, 1),
D(1, –1) ÁËÊ 1 1 ¯˜ˆ
2 2
Let P be (0, 1) and Q at ( 2, 0) T1 , is the vertex of the parabola.

3.80 Coordinate Geometry Booster

AT1 = 1 and T2T3 = latus rectum of parabola Let its roots are m1, m2.
2 Therefore, m1m2 = –1
fi the tangents are mutually perpendicular.
=4¥ 1 =2 2
2 As we know that the point of intersection of two mutu-

Thus, the area of the DT1T2T3 ally perpendicular tangents is the director circle.
= 1 ¥ 1 ¥ 2 2 = 1 Sq unit. So, the equation of the director circle is
22
x2 + y2 = 338
Note. No questions asked in 2015
Therefore, the Statement II is the correct explanation of
80. Given AB || CD ◊ CD = 2AB.
the Statement I.
D RC 82. Given circle is (x + 3)2 + (y – 5)2 = 4

So, the radius = 2

Distance between the parallel lines L1 and L2 is

rQ ( p + 3) - ( p - 3) = 6 < radius (2)
SO 4+9 13

So, the Statement II is false, but the Statement I is true.

AP B Comprehension

Let AB = 2a, CD = a YQ

and the radius of the circle be r.

Let the circle touches AB at P, BC at Q, AD at R E D 33 , 3
2 2
and CD at S.
C
Then AR = AP = r, BP = BQ = a – r, 1 3x + y = 6
F PX
DR = DS = r and CQ = CS = 2a – r X¢ R

In DBEC, Y¢
BC2 = BE2 + EC2
83. Co-ordinates of C are
fi (a – r + 2a – r)2 = (2r)2 + a2
fi (3a – 2r)2 = (2r)2 + a2 x-3 3 y-3
fi 9a2 + 4r2 – 12ar = 4r2 +a2 2 2
fi a= 3r fi = = -1
ÊÁË p ˆ¯˜ sin ÊÁË p ˆ˜¯
2 cos 6 6

Also, ar(Quad. ABCD) = 18 fi x-3 3 =- 3, y- 3 =-1
fi ar(Quad. ABED) + ar(DBCE) = 18 2 2 22
fi a ◊ 2r + 1 ◊ a ◊ 2r = 18
fi x = 3, y = 1
2
fi 3ar = 18 Thus, C = ( 3, 1)
fi 3 ¥ 3 ¥ r2 = 18 Hence, the equation of the circle is

2 (x - 3)2 + ( y - 1)2 = 1
fi r2 = 4 84. Clearly, the point F is ( 3, 0).
fi r=2
Thus, the radius is r = 2. Now the co-ordinates of E are
81. The equation od any tangent to the given circle is
fi x — 3 = y -1 =1
y = mx + a 1 + m2 cos (150°) sin (150°)

which is passing through (17, 7). fi x— 3 = y -1 =1
-3 1
Thus, 7 = 17m + 13 1 + m2 2
2
fi (7 – 17m)2 = 169(1 + m2)
fi 49 + 289 m2 – 238m = 169(1 + m2) fi x — 3 = - 3, y =1- 1
fi 120m2 – 238m – 120 = 0 22

Circle 3.81

fi x= 3- 3, y= 1 Similarly, OM 2 = 2 cos ËÁÊ p ¯˜ˆ
22 k
It is given that,
fi x= 3, y= 1
22 OM1 + OM2 = 2

Ê 3 1ˆ fi 2 cos ÊËÁ p ˜ˆ¯ + 2 cos ÊËÁ p ˜ˆ¯ = ( 3 + 1)
Therefore, E = ÁË 2 2˜¯ 2k k
, , F = ( 3, 0)

fi cos ËÁÊ p ˜¯ˆ + cos ËÁÊ p ¯˜ˆ = Ê 3 + 1ˆ
2k k ËÁ 2 ˜¯
85. Co-ordinates of Q are ( 3, 3).
fi cos Ê q ˆ + cos (q ) = Ê 3 + 1ˆ Êpˆ
The equation of QR is y - 0 = 3 (x - 0) ËÁ 2 ˜¯ ÁË 2 ˜¯ ËÁ k ˜¯
3 where =q

fi y = 3x fi cos Ê q ˆ + 2 cos2 Ê q ˆ -1= Ê 3 + 1ˆ
ÁË 2 ˜¯ ÁË 2 ¯˜ ËÁ 2 ¯˜
and the equation of RP is y = 0.
86. Given circle is x2 + y2 – 6x – 4y – 11 = 0

B fi 2 cos2 Êqˆ + cos Ê q ˆ - Ê 3+ 3ˆ = 0
ÁË 2˜¯ ËÁ 2 ˜¯ ÁË 2 ¯˜

CP fi 2b2 + b - Ê 3+ 3ˆ = 0, b = cos Ê qˆ
ÁË 2 ˜¯ ÁË 2 ˜¯

A fi 4b2 + 2b - ( 3 + 3) = 0

Therefore, the centre is (3, 2) fi b = —2 ± 4 + 16(3 + 3)
Since CA and CB are perpendicular to PA and PB. 8
So, CP is the diameter of the circumcircle PAB.
Thus, the equation of the circumcircle triangle PAB is fi b = —1 ± 1 + 4(3 + 3)
2
(x – 3)(x – 1) + (y – 2)(y – 8) = 0
fi x2 + y2 – 4x – 10y + 19 = 0 fi b = —1 ± 1 + 4 3 + 12
87. Clearly, the length of the perpendicular from the centre 2
of the circle is equal to the radius of the circle.
fi b = —1 ± (1 + 2 3)2 = —1 ± (1 + 2 3)
Thus, h◊0 + k ◊0 - 1 = 2 22
h2 + k2
fi b= 3, - 3 -1
fi (h2 + k2) = 1 22
4
fi b= 3
Hence, the locus of the (h, k) is 2
x2 + y2 = 1
4 fi cos Êqˆ = 3 = cos ÁËÊ p ˜¯ˆ
ËÁ 2¯˜ 2 6
88. D M2 C

O fi cos ËÁÊ q ¯ˆ˜ = cos ÊÁË p ¯ˆ˜
2 6

A M1 B fi q=p
3
In DOM1A,
fi p =p
OA = 2, –AOM1 = p k3
2k

cos ÊÁË p ˆ˜¯ = OM1 fi OM1 = 2 cos ÊÁË p ˆ˜¯ fi k=3
2k 2 2k
fi [k] = 3

3.82 Coordinate Geometry Booster
89. Let L: 2x – 3y = 1
91. (i) Equation of the tangent to the circle x2 + y2 = 4
Y at P( 3, 1) is 3x + y = 4
i.e PT: 3x + y = 4
X¢ O L : 2x – 3y = 1 X Now, m(PT ) = - 3
So, m(L) = 1 .
Y¢ 3
The line L is y = m(x - 3) ± a 1 + m2
and S: x2 + y2 £ 6
fi y = 1 (x - 3) ± 1 1 + 1
If L > 0 and S < 0, the points lie on the smaller part. 33

Thus, the points Ê 2, 3ˆ and Ê 1 , - 1 ˆ lie inside the tri- fi y = 1 (x - 3) ± 2
ËÁ 4˜¯ ÁË 4 4 ˜¯ 33

angle. fi y = x -5, x -1
33
90. The equation of the chord bisected at P(h, k) is
fi x - 3y - 5 = 0, x - 3y - 1 = 0
hx + ky = h2 + k2 …(i) (ii) Given circles are

Y x2 + y2 = 4, (x – 3)2 + y2 = 1

M

2

C1(0,0) C2(3,0) P(6, 0)

X¢ M

O P(h, k) B Clearly, the point of intersection is (6, 0).
The equation of the direct common tangent is
A 4a
5 y – 0 = m(x – 6)
L Q a , –4 fi mx – y + 6m = 0

Y¢ Now, C1M = 2

Let any point on line LM be ÁÊË a , 4a - 4ˆ˜¯ . fi 6m = 2
5 m2 + 1

The equation of the chord of contact is fi m2 + 1 = 9m2
fi 8m2 = 1
ax + Ê 4a - 4¯˜ˆ y = 9 … (ii) fi m=± 1 .
ËÁ 5
22
Comparing Eqs (i) and (ii), we get Hence, the equation of the common tangents are

h = k = h2 + k2 y = ± 1 (x - 6) and x = 2
a 4a - 9 22
4
5 92. Given circles are x2 + y2 – 2x – 15 = 0
and x2 + y2 = 1
fi a = 20h Let the equation of the circle is
4h - 5k x2 + y2 + 2gx + 2fy + c = 0
It passes through (0, 1), so
Therefore, h = h2 + k2 1 + 2f + c = 0
20h 9 Applying condition of orthogonality, we have
–2g = c – 15, 0 = c – 1
4h - 5k fi c = 1, g = 7, f = –1

fi 4h - 5k = h2 + k 2 Thus, r = 49 + 1 — 1 = 7
20 9
and centre = (–7, 1)
fi 20(h2 + k2) = 9(4h – 5)

Hence, the locus of P(h, k) is
20(x2 + y2) = 9(4x – 5y)

Circle 3.83
93.
Thus, the area of the equadrilateral PQRS
Y R(2, 4) = 1 ¥ (2 + 8) ¥ 3 = 15
P(–1, 1) 2

94. Ans. (a, c)

X¢ O X
Q(–1, –1)

S(2, –4) E = Ê a q , a ˆ
Y¢ ÁË tan ˜¯

The equation of any tangent to the parabola can be con- cos q + a sin q = 1
sidered as y = mx + a = mx + 2 .
a = tan ÊÁË q ¯ˆ˜
mm 2
i.e. m2x – my + 2 = 0
As we know that the length of the perpendicular from Hence, the locus is y2 = 1 – 2x.
the centre to the tangent to the circle is equal to the
radius of a circle. Thus, the required points are Ê 1 , 1ˆ and Ê 1, - 1ˆ .
Thus, 2 = 2 ËÁ 3 3 ¯˜ ÁË 3 3 ¯˜

m4 + m2

m4 + m2 = 2
m4 + m2 – 2 = 0
(m2 + 2)(m2 – 1) = 0
m = ±1
Hence, the equation of the tangents are

y = x + 2, y = –x – 2

Therefore, the points P, Q are (–1, 1), (–1, –1) and R, S
are (2, 4) and (2, –4) respectively.



CHAPTER Parabola 4.1

4 Parabola

CONCEPT BOOSTER (ii) Parabola

1. INTRODUCTION The section of a right circular cone by a plane which is paral-
lel to a generator of a cone is called a parabola.
It is believed that the first definition of a conic section is due
to Menaechmus (died 320 BC). His work did not survive and Generator
is only known through secondary accounts. The definition
used at that time differs from the one commonly used today. Parabola
It requires the plane cutting the cone to be perpendicular to
one of the lines (a generatrix), that generates the cone as a Plane
surface of revolution. Thus the shape of the conic is deter-
mined by the angle formed at the vertex of the cone (between (iii) Ellipse
two opposite generatrices). If the angle is acute, the conic is
an ellipse; if the angle is right, the conic is a parabola; and if The section of a right circular cone by a plane which is nei-
the angle is obtuse, the conic is a hyperbola. ther parallel to a generator of a cone nor parallel or perpen-
dicular to the axis of a cone is called an ellipse.
Note: The circle cannot be defined in this way and was not
considered as a conic at this time. Ellipse

Euclid (300 BC) is said to have written four books on con-
ics but these were lost as well. Archimedes (died 212 BC)
is known to have studied conics, having determined the area
bounded by a parabola and an ellipse. The only part of this
work to survive is a book on the solids of revolution of conics.

2. BASIC DEFINITIONS

(i) Circle

The section of a right circular cone by a plane which is paral-
lel to its base is called a circle.

Plane

Circle (iv) Hyperbola

The section of a double right circular cone by a plane which
is parallel to the axis of a cone is called a hyperbola.

Plane

4.2 Coordinate Geometry Booster

Hyperbola Latus rectum
Any chord passing through the focus and perpendicular to the
axis is known as latus rectum of the conic section.

Centre
The point which bisects every chord of the conic passing
through it, is called the centre of the conic section.

4. RECOGNITION OF CONICS

3. CONIC SECTION A general equation of 2nd degree is
ax2 + 2hxy + by2 + 2gx + 2fy + c = 0,
It is the locus of a point which moves in a plane in such a way ahg
that its distance from a fixed point to its perpendicular dis-
tance from a fixed straight line is always constant. The fixed where D = h b f
point is called the focus of the conic and this fixed straight gf c
line is called the directrix of the conic and this constant ratio
is known as the eccentricity of the conic. It is denoted as e. = abc + 2fgh – af2 – bg2 – ch2

MP and a h = ab - h2
H=
hb

S(Focus) There are two types of conics.
(i) Degenerate conic, and
Directirix (ii) Non-degenerate conic.
We use the term degenerate conic sections to describe the
Conic Section with Respect to Eccentricity single point, single line and pair of lines and the term non-
degenerate conic sections to describe those conic sections
(i) If e = 0, the conic section is called a circle that are circles, parabolas, ellipses or hyperbolas.
(ii) If e = 1, the conic section is called a parabola. A non-degenerate conic represents
(iii) If e < 1, the conic section is called an ellipse. (i) a circle if, D π 0, h = 0, a = b
(iv) If e > 1, the conic section is called a hyperbola. (ii) a parabola if D π 0, H = 0
(v) If e = 2 , the conic section is called a rectangular hy- (iii) an ellipse, if D π 0, H < 0
(iv) a hyperbola, if D π 0, H > 0
perbola. (v) a rectangular hyperbola, if D π 0, H > 0 and a + b = 0.
Now, the centre of the conics is obtained by
Some Important Definitions to Remember
d f = 2ax + 2hy + 2g = 0
Axis: The straight line passing through the focus and perpen- dx
dicular to the directrix is called the axis of the conic section.
and d f = 2hx + 2by + 2f = 0 .
Vertex: The point of intersection of the conic section and the dy
axis is called the vertex of the conic section.
fi ax + hy + g = 0, hx + by + f = 0
Double ordinate
Any chord, which is perpendicular to the axis of the conic Solving the above equations, we get the required centre of
section, is called a double ordinate of the conic section. the given conic.

Focal chord 5. EQUATION OF CONIC SECTION
Any chord passing through the focus is called the focal chord
of the conic section. MP

Focal distance S(Focus)
The distance between the focus and any point on the conic is
known as the focal distance of the conic section. Directirix

Parabola 4.3

Let the focus be (h, k), directrix be ax + by + c = 0 and the Definition 3
In algebra, the parabolas are frequently encountered as
eccentricity is e. SP =e graphs of quadratic functions, such as y = ax2 + bx + c or
Then the equation of the conic section is x = ay2 + by + c.
PM
Definition 4
fi (x - h)2 + ( y - k)2 = e ax + by + c It is a section of a conic, whose eccentricity is 1.
a2 + b2
Definition 5
fi (x - h)2 + ( y - k )2 = e2 Ê (ax + by + c)2 ˆ A plane curve formed by the intersection of a right circular
ËÁ a2 + b2 ˜¯ cone and a plane parallel to an element of the cone is called
parabola.
which is the general equation of the conic section.

6. PARABOLA Generator

The term parabola comes from Greek word, para ‘alongside, Parabola
nearby, right up to,’ and bola, from the verb ballein means
‘to cast, to throw.’ Understandably, parallel and many of its
derivatives start with the same root. The word parabola may
thus mean ‘thrown parallel’ in accordance with the definition.

7. MATHEMATICAL DEFINITIONS Plane

Definition 1 Definition 6
It is the locus of a point which moves in a plane in such a A conic ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represents a
way that its distance from a fixed point is equal to its distance parabola if
from a fixed straight line. The fixed point S is called the focus
and the fixed straight line OM is called the directrix. (i) D π 0
(ii) h2 – ab = 0, where
Y
P D = abc + 2fgh – af 2 – bg2 – ch2

M

8. STANDARD EQUATION OF A PARABOLA

OS X Y
M
P

X¢ ZO S X

Definition 2 Y¢
Let D be a line in the plane and F a fixed point not on D. A
parabola is the collection of points in the plane that are equi- Let S be the focus and MN be the directrix of the parabola.
distant from F and D. The point F is called the focus and the Draw SZ perpendicular to ZM and let O be the mid-point
line D is called the directrix.
of FN.
Y Thus OS = OZ
So O lies on the parabola.
D Consider O as the origin and OX and OY as x and y axes,

O FX respectively.
Let OS = OZ = a.
Then the co-ordinates of F is (a, 0) and the equation of

ZM is x + a = 0.
Now from the definition of the parabola, we get,
SP = PM
fi SP2 = PM2

4.4 Coordinate Geometry Booster

fi (x – a)2 + (y – 0)2 = (x + a)2 (vi) Length of the latus rectum: 4a
fi y2 = 4ax (vii) Extremities of the latus rectum are: L(a, 2a), L¢(a, –2a)
(viii) Equation of the latus rectum is: x = a
which is the required equation of a parabola. This is also (ix) Parametric equations of the parabola:
known as horizontal parabola or right-ward parabola.
y2 = 4ax are x = at2 and y = 2at
9. IMPORTANT TERMS RELATED TO PARABOLA
(x) Focal distance: x + a
Y (xi) Any point on the parabola can be considered as
M
(at2, 2at).

P Parabola openning leftwards
i.e y2 = –4ax
X¢ Z O S X
Y
L(–a, 2a)

P(x, y) M

Y¢ X¢ S(–a, 0) O X

Focus L¢(–a, –2a) x–a=0
It is the fixed point with reference to which the parabola is
constructed. Here, S is the focus. Y¢

Directrix (i) Vertex is (0, 0)
It is a straight line outside the parabola. Here ZM is the
directrix. (ii) Focus is (–a, 0)

Axis of symmetry (iii) Equation of the directrix is x – a = 0
It is the line which is perpendicular to the directrix and passes
through the focus. It divides the parabola into two equal halves. (iv) Equation of the axis is y = 0

Vertex (v) Equation of the tangent at the vertex is x = 0
It is the point on the axis of symmetry that intersects the pa-
rabola when the turn of the parabola is the sharpest. (vi) Length of the latus rectum is 4a

The vertex is halfway between the directrix and the focus. (vii) Extremities of the latus rectum are:

Focal chord L(–a, 2a), L¢(a, –2a)
It is any chord that passes through the focus.
(viii) Equation of the latus rectum is x = –a
Latus rectum
It is that focal chord which is perpendicular to the axis of (ix) Parametric equations of the parabola:
symmetry. The latus rectum is parallel to the directrix. Half
of the latus rectum is called the semi-latus rectum. y2 = –4ax are x = –at2 and y = 2at

Focal parameter (x) Focal distance is x – a
The distance from the focus to the directrix is called the focal
parameter. (xi) Any point on the parabola can be considered as
(–at2, 2at).
Focal distance
The distance between any point on the parabola to the focus Parabola opening upwards
is called the focal distance. Here, SP is the focal distance. i.e. x2 = 4ay

Parametric equation Y
From the equation of the parabola, we can write
S P
y = 2x = t (0, a)
2a y
Then x = at2, y = 2at, where t is a parameter. X¢ O X
The equations x = at2 and y = 2at are called the parametric
equations and the point (at2, 2at) is also referred to as the y+a=0 Z M
point t.
(i) Vertex is: (0, 0) Y¢
(ii) Focus is: (a, 0)
(iii) Equation of the directrix is: x + a = 0. (i) Vertex is (0, 0)
(iv) Equation of the axis is: y = 0
(v) Equation of the tangent at the vertex is: x = 0 (ii) Focus is (0, a)

(iii) Equation of the directrix is y + a = 0

(iv) Equation of the axis is x = 0

(v) Equation of the tangent at the vertex is y = 0

(vi) Length of the latus rectum 4a
(vii) Extremities of the latus rectum are L(2a, a), L¢(–2a, a)

(viii) Equation of the latus rectum is y = a
(ix) Parametric equations of the parabola x2 = 4ay are

x = –2at and y = at2

Parabola 4.5

(x) Focal distance is y + a 11. EQUATION OF A PARABOLA WHEN THE VERTEX IS
(h, k) AND AXIS IS PARALLEL TO x-AXIS
(xi) Any point on the parabola can be considered as
(2at, at2). The equation of the parabola y2 = 4ax can be written as
(y – 0)2 = 4a(x – 0)
Parabola Opening Downwards
i.e. x2 = –4ay The vertex of the parabola is O(0, 0). Now the origin is
shifted to V(h, k) without changing the direction of axes, its
Y equation becomes (y – k)2 = 4a(x – h)

ZM Y

X¢ O X

P

S

Y¢ V(h, k) y=k
O X
(i) Vertex is (0, 0)

(ii) Focus is (0, –a)

(iii) Equation of the directrix is y – a = 0

(iv) Equation of the axis is x = 0 Thus its focus is F(a + h, k), latus rectum = 4a and the
equation of the directrix is
(v) Equation of the tangent at the vertex is y = 0
x = h – a, i.e. x + a – h = 0
(vi) Length of the latus rectum is 4a The parametric equation of the curve (y – k)2 = 4a(x – h)
are x – h + at2 and y = k + 2at.
(vii) Extremities of the latus rectum are:
12. EQUATION OF A PARABOLA WHEN THE VERTEX IS
L(2a, –a), L¢(–2a, –a) (h, k) AND AXIS IS PARALLEL TO y-AXIS

(viii) Equation of the latus rectum is y = –a Y
(ix) parametric equations of the parabola x2 = –4ay are

x = 2at and y = –at2

(x) Any point on the parabola can be considered as
(2at, –at2).

(xi) Focal distance is y – a

10. GENERAL EQUATION OF A PARABOLA V(h, k) y=k
X¢ O X

Y P(x, y) Y¢ x = h

M S(h, k) The equation of a parabola with the vertex V(h, k) is
(x – h)2 = 4a(y – k)
X¢ Z X
O Thus, its focus is F(h, a + k), latus rectum = 4a and the

equation of the directrix is

y = k – a, i.e. y + a – k = 0
The parametric equation of the curve (x – h)2 = 4a(y – k)
are x = h + 2at and y = k + at2.

Y¢ Note: The equation of a parabola, whose axis is parallel to
y-axis can also be considered as y = ax2 + bx + c.
Let S(h, k) be the focus and lx + my + n = 0 is the equation of
the directrix and P(x, y) be any point on the parabola. Polar form of a Parabola

Then, In polar coordinates, the equation of a parabola with param-
SP = PM eters r and q and the centre (0, 0) is given by

fi (x - h)2 + ( y - k)2 = lx + my + n r = - 2a
(l2 + m2 ) 1 + cos q

fi (x - h)2 + (y - k)2 = (lx + my + n)2 13. FOCAL CHORD
(l2 + m2 )
Any line passing through the focus and intersects the parabola
fi m2x2 + l2y2 – 2lmxy + (term)x in two distinct points, it is known as focal chord of the parabola.

+ (term)y + (constant term) = 0 Any point on the parabola y2 = 4ax can be considered as
fi (mx – ly)2 + 2gx + 2fy + c = 0 (at2, 2at).

which is the general equation of a parabola.

4.6 Coordinate Geometry Booster

14. POSITION OF A POINT RELATIVE TO A PARABOLA Equation of the Tangent to a Parabola in Different
Forms
Consider the parabola y2 = 4ax and the point be (x1, y1).
(i) Point form
Y P(x1, y1) The equation of the tangent to the parabola y2 = 4ax at
(x1, y1) is
L X
OM yy1 = 2a(x + x1)

Now the given equation is

y2 = 4ax.

The point (x1, y1) lies outside, on and inside of the pa- Differentiating with respect to x, we get
rabola y2 = 4axy2 = 4ax according as 2y dy = 4a
dx
y12 – 4ax1 > 0, = 0, < 0
fi dy = 2a
15. INTERSECTION OF A LINE AND A PARABOLA dx y

Let the parabola be y2 = 4ax Now m = Ê dy ˆ = 2a .
and the line be y = mx + c. ËÁ dx ¯˜ (x1, y1) y1
MT
Eliminating x between R Thus the equation of tangent is
these two equations, we get fi yy1 – y12 = 2ax – 2ax1
fi yy1 = 2ax – 2ax1 + 4ax1
y2 = 4a ËÁÊ y - c ¯ˆ˜ P
m = 2a (x + x1)

fi my2 – 4ay + 4ac = 0 Q which is the required equation of the tangent to the pa-
rabola y2 = 4ax at (x1, y1).
N
(ii) Parametric form
The given line will cut the The equation of the tangent to the parabola y2 = 4ax at
(at2, 2at) is
parabola in two distinct, co-
yt = x + at2
incident and imaginary points according as
The equation of the tangent to the parabola y2 = 4ax at
D > 0, = 0, < 0
fi 16a2 – 16amc > 0, = 0, < 0 (x1, y1) is

fi a > cm, a = cm, a < cm yy1 = 2a(x + x1)
fi y ◊ 2at = 2a(x + at2)
Condition of tangency: The line y = mx + c will be a tangent
to the parabola y2 = 4ax, if c = a . fi yt = x + at2

m (iii) Slope form
The equation of any tangent to the parabola can be consid- The equation of the tangent to the parabola y2 = 4ax at
ered as y = mx + a .
Ê a , 2a ˆ is
m ÁË m2 m ˜¯

The co-ordinates of the point of contact in terms of m is y = mx + a , where m = slope
m
Ê a , 2a ˆ
ÁË m2 m ˜¯

The equation of the tangent to the parabola y2 = 4ax at

16. TANGENT (x1, y1) is

If a line intersects the parabola in two coincident points, it is yy1 = 2a(x + x1) …(i)
known as the tangent to a parabola.
Here, m = 2a fi y1 = 2a
y1 m

Q Since the point (x1, y1) lies on the parabola y2 = 4ax, we
have,
Q1
y12 = 4ax1.
Q2
fi 4ax1 = 4a2
Q3 m2
Q4
PT fi x1 = a
m2

Parabola 4.7

Putting the values of x1 and y1 in Eq. (i), we get 17. NORMAL

y ◊ ÁËÊ 2aˆ = 2a Ê x + a ˆ It is a line which is perpendicular to the point of contact to
m ˜¯ ËÁ m2 ˜¯ the tangent.

fi y = mx + a N
m

(iv) Condition of tangency P
The line y = mx + c will be a tangent to the parabola
y2 = 4ax is c = a .
m

Note: Any tangent to the parabola can be considered T
as y = mx + a .
Here PT is a tangent and PN is a normal.
m
Equation of Normals to the Parabola in Different
(v) Director circle Forms
The locus of the point of intersection of two perpen-
dicular tangents to a parabola is known as the director (i) Point form
circle. The equation of the normal to the parabola y2 = 4ax at

(vi) The equation of the pair of tangents can be drawn to the (x1, y1) is
parabola from the point (x1, y1). y - y1 = x - x1
y1 2a
M(h, k)
The equation of the tangent to the parabola y2 = 4ax at

(x1, y1) is …(i)
yy1 = 2a(x + x1)

P(x1, y1) Slope of the tangent is m(T ) = 2a
y1

N Slope of the normal is m(N ) = - y1
2a

Thus the equation of the normal is

Let (h, k) be any point on either of the tangents drawn y - y1 = - y1 (x - x1)
2a
from (x1, y1).

The equation of the line joining (x1, y1) and (h, k) is (ii) Parametric form

y - y1 = k - y1 (x - x1) The equation of the normal to the parabola
h - x1
y2 = 4ax at (at2, 2at) is
fi y = k - y1 x + hy1 - kx1 y = –tx + 2at + at3

h - x1 h - x1 a. As we know that the equation of the normal to the pa-
rabola y2 = 4ax at (x1, y1) is
If this be a tangent, it must be of the form y = mx + m
y - y1 = - x - x1
Thus, m = k - y1 and a = hy1 - kx1 y1 2a
h - x1 m h - x1
Replacing x1 by at2 and y1 by 2at, we get
Therefore, by multiplication we get
y - 2at = - x - at2
a = Ê k - y1 ˆ Ê hy1 - kx1 ˆ 2at 2a
ÁË h - x1 ¯˜ ËÁ h- x1 ¯˜
fi y = –tx + 2at + at3

fi a(h – x1)2 = (k – y1)(hy1 – kx1) which is the required equation of the normal to the giv-

Hence, the locus of the point (h, k) is en parabola.

fi a(x – x1)2 = (y – y1)(xy1 – yx1) (iii) Slope form
fi (y2 – 4ax)(y12 – 4ax1) = {yy1 – 2a(x + x1)}2.
fi SS1 = T2 The equation of the normal to the parabola
where, S: (y2 – 4ax), S1:(y12 – 4ax1) y2 = 4ax at (ax2, –2am) is
y = mx – 2am – am3
and T:{yy1 – 2a(x + x1)}.
As we know that the equation of the normal to the pa-

rabola

4.8 Coordinate Geometry Booster
y2 = 4ax is
y2 = 4ax at (at2, 2at) is …(i) yy1 = 2a(x + x1)
y = –tx + 2at + at3
Q
The slope of the normal is
P Chord of contact
m = –t
R
fi t = –m
19. CHORD BISECTED AT A GIVEN POINT
Putting the values of m in Eq. (i), we get
y = mx – 2am – am3 Y
A
which is the required equation of the normal to the pa-
rabola y2 = 4ax at (am2, –2am).

(iv) Condition of normal

The line y = mx + c will be a normal to the parabola
y2 = 4ax, if c = –2am – am3 and the co-ordinates of the
point of contact are (am2, –2am).

(v) Co-normal points

In general, three normals can be drawn from a point to

a parabola and their feet (points) where they meet the

parabola are called the co-normal points.

Y OM X

A P(h, k) B

O X The equation of the chord of the parabola
C y2 = 4ax is bisected at the point (x1, y1) is
B T = S1

Here A, B and C are three co-normal points. fi yy1 – 2a(x + x1) = y12 – 4ax1
Let P(h, k) be any given point y2 = 4ax be a parabola. where
The equation of any normal to the parabola
T: y1 – 2a(x + x1), S:y12 – 4ax1
y2 = 4ax is
y = mx – 2am – am3 20. DIAMETER
which passes through P(h, k). Then
k = mh – 2am – am3 The locus of the mid-points of a system of parallel chords to a
fi am3 + (2a – h)m + k = 0 parabola is known as the diameter of the parabola.
which is a cubic equation in m. So it has three roots.
Thus, in total, three normals can be drawn from a point Y
lies either outside or inside of a parabola.
M(h, k) y = 2a/m
Notes:
1. We can draw one and only one normal to a pa- X
rabola, if a point lies on the parabola. O
2. From an external point to a parabola, only one
normal can be drawn. The equation of the diameter to the parabola y2 = 4ax
bisecting a system of parallel chords with slope m is
18. CHORD OF CONTACT
y = 2a
The chord joining the points of contact of two tangents drawn m
from an external point to a parabola is known as the chord of
contact. Let (h, k) be the mid-point of the chord y = mx + c of the
parabola y2 – 2ah.
The equation of the chord of contact of tangents drawn
from a point (x1, y1) to the parabola Then, T = S1
fi ky – 2a(x + h) = k2 – 2ah

Parabola 4.9

Now slope = 2a 21. REFLECTION PROPERTY OF A PARABOLA
k
All rays of light coming from the positive direction of x-axis
m = 2a fi k = 2a and parallel to the axis of the parabola are reflected through
km the focus of the parabola

Hence the locus of the mid-point (h, k) is y = 2a . Y
m MP

Note: Any line which is parallel to the axis of the parabola O S X
drawn through any point on the parabola is called the di- Z
ameter of the parabola and its equation is the y-coordinate
of that point.

EXERCISES

LEVEL I (Problems based on 12. Find the equation of a parabola whose vertex is (1, 2)
Fundamentals) and the axis is parallel to x-axis and also passes through
the point (3, 4).
1. What conic does ax + by = 1 represent?
13. If the axis of a parabola is parallel to y-axis, the ver-
2. If the conic x2 – 4xy + ly2 + 2x + 4y + 10 = 0 represents tex and the length of the latus rectum are (3, 2) and 12
a parabola, find the value of l. respectively, find its equation.

3. If the conic 16(x2 + ( y - 1)2 ) = (x + 3y - 5)2 repre- PROPERTIES OF THE FOCAL CHORD
sents a non-degenerate conic, write its name and also
find its eccentricity. 14. If the chord joining P(at12, 2at1) and is the focal chord,
prove that t1t2 = –1.
4. If the focus and the directrix of a conic be (1, 2) and
x + 3y + 10 = 0 respectively and the eccentricity be 15. If the point (at2, 2at) be the extremity of a focal chord
1 , then find its equation.
2 of the parabola y2 = 4ax, prove that the length of the

5. Find the equation of a parabola, whose focus is (1, 1) Ê 1ˆ 2
and the directrix is x – y + 3 = 0. ËÁ t ¯˜
focal chord is a t + .

16. If the length of the focal chord makes an angle q with

the positive direction of x-axis, prove that its length is
4a cosec2q.
17. Prove that the semi-latus rectum of a parabola y2 = 4ax

ABC OF PARABOLA is the harmonic mean between the segments of any

6. Find the vertex, the focus, the latus rectum, the direc- focal chord of the parabola.

trix and the axis of the parabolas 18. Prove that the length of a focal chord of the parabola
(i) y2 = x + 2y + 2
(ii) y2 = 3x + 4y + 2 varies inversely as the square of its distance from the
(iii) x2 = y + 4x + 2
(iv) x2 + x + y = 0 vertex.
7. If the focal distance on a point to a parabola y2 = 12x is
6, find the co-ordinates of that point. 19. Prove that the circle described on the focal chord as the

8. Find the equation of a parabola, whose focus (–6, –6) diameter touches the tangent to the parabola.

and the vertex is (–2, –2). 20. Prove that the circle described on the focal chord as the
9. The parametric equation of a parabola is x = t 2 + 1 and
diameter touches the directrix of the parabola.
y = 2t + 1. Find its directrix.
10. If the vertex of a parabola be (–3, 0) and the directrix POSITION OF A POINT RELATIVE TO A PARABOLA

is x + 5 = 0, find its equation. 21. If a point (l, –l) lies in an interior point of the parabola
11 Find the equation of the parabola whose axis is parallel y2 = 4x, find the range of l.

to y-axis and which passes through the points (0, 2), 22. If a point (l, 2) is an exterior point of both the parabo-
(–1, 0) and (1, 6). las y2 = (x + 1) and y2 = –x + 1, find the value of l.

INTERSECTION OF A LINE AND A PARABOLA

23. If 2x + 3y + 5 = 0 is a tangent to the parabola y2 = 8x,
find the co-ordinates of the point of contact.